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A first-course textbook, not Phil's own writing, by Marcel B. Finan, dated August 2009. It reviews calculus and ODE background, then covers first-order PDEs and characteristics, the wave, heat and Laplace equations, Fourier series, separation of variables, and Laplace and Fourier transform methods. Chapters include practice problems and an answers section.
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Undergraduate Notes in Mathematics
Arkansas Tech University
Department of Mathematics
A First Course of Partial Di erential
Equations
in Physical Sciences and Engineering
Marcel B. Finan
Arkansas Tech University
c
All Rights Reserved
2
Preface
Partial di erential equations are often used to construct models of the most
basic theories underlying physics and engineering. The goal of this book is to
develop the most basic ideas from the theory of partial di erential equations,
and apply them to the simplest models arising from the above mentioned
elds.
It is not easy to master the theory of partial di erential equations. Unlike
the theory of ordinary di erential equations, which relies on the fundamental
existence and uniqueness theorem, there is no single theorem which is central
to the subject. Instead, there are separate theories used for each of the major
types of partial di erential equations that commonly arise.
It is worth pointing out that the preponderance of di erential equations aris-
ing in applications, in science, in engineering, and within mathematics itself,
are of either rst or second order, with the latter being by far the most preva-
lent. We will mainly cover these two classes of PDEs.
This book is intended for a rst course in partial di erential equations at
the advanced undergraduate level for students in engineering and physical
sciences. It is assumed that the student has had the standard three semester
calculus sequence, and a course in ordinary di erential equations.
Marcel B Finan
August 2009
3
4 PREFACE
Contents
Preface 3
Preliminaries 7
1 Some Results of Calculus . . . . . . . . . . . . . . . . . . . . . . . 7
2 Sequences of Functions: Pointwise and Uniform Convergence . . . 14
Review of Some ODEs Results 23
3 The Method of Integrating Factor . . . . . . . . . . . . . . . . . . 23
4 The Method of Separation of Variables for ODEs . . . . . . . . . 28
5 Second Order Linear ODEs . . . . . . . . . . . . . . . . . . . . . 33
Introduction to PDEs 43
6 The Basic Concepts . . . . . . . . . . . . . . . . . . . . . . . . . 43
7 Solutions and Related Topics . . . . . . . . . . . . . . . . . . . . 52
First Order Partial Di erential Equations 65
8 Classi cation of First Order PDEs . . . . . . . . . . . . . . . . . 65
9 The One Dimensional Spatial Transport Equations . . . . . . . . 71
10 The Method of Characteristics . . . . . . . . . . . . . . . . . . . 79
11 The Cauchy Problem for First Order Quasilinear Equations . . . 86
Second Order Linear Partial Di erential Equations 99
12 Second Order PDEs in Two Variables . . . . . . . . . . . . . . . 99
13 Hyperbolic Type: The Wave equation . . . . . . . . . . . . . . . 104
14 Parabolic Type: The Heat Equation in One-Dimensional Space . 112
15 An Introduction to Fourier Series . . . . . . . . . . . . . . . . . 120
16 Fourier Sines Series and Fourier Cosines Series . . . . . . . . . . 132
17 Separation of Variables for PDEs . . . . . . . . . . . . . . . . . 139
5
6 CONTENTS
18 Solutions of the Heat Equation by the Separation of Variables
Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
19 Elliptic Type: Laplace's Equations in Rectangular Domains . . . 154
20 Laplace's Equations in Circular Regions . . . . . . . . . . . . . . 165
The Laplace Transform Solutions for PDEs 177
21 Essentials of the Laplace Transform . . . . . . . . . . . . . . . . 177
22 Solving PDEs Using Laplace Transform . . . . . . . . . . . . . . 192
The Fourier Transform Solutions for PDEs 199
23 Complex Version of Fourier Series . . . . . . . . . . . . . . . . . 199
24 The One Dimensional Fourier Transform . . . . . . . . . . . . . 205
25 Applications of Fourier Transforms to PDEs . . . . . . . . . . . 213
Answers and Solutions 221
Preliminaries
In this chapter we include some results from calculus which we will use often
in the study of partial di erential equations. Details and proof of these results
can be found in most calculus books.
1 Some Results of Calculus
The rst result provides a mean of showing when a function is zero on an
interval.
Theorem 1.1
(a) Suppose that fis continuous on an interval IRsuch thatRb
af(x)dx= 0
for all subintervals [ a;b]I:Thenf(x) = 0 for all x2I:
(b) Suppose that f: [a;b]!Ris continuous and non-negative. IfRb
af(x)dx=
0 thenf(x) = 0 on [a;b]:
(c) Suppose that f: [a;b]!Ris continuous such thatRb
af(x)g(x)dx= 0
for all continuous functions gon [a;b]:Thenf(x) = 0 on [a;b]:
Proof.
(a) Fixa2I:Letx2I:By the Fundamental Theorem of Calculus we have
0 =d
dxZx
af(t)dt=f(x):
Sincexwas arbitrary, we have f(x) = 0 for all x2I:
(b) Suppose the contrary. That is, suppose that x02[a;b] such that f(x0)>
0:By the continuity of f(x) atx0;there is a >0 such thatjx x0j<
impliesjf(x) f(x0)j<f(x0)
2:That is,jx x0j<impliesf(x)>f(x0)
2>0:
7
8 PRELIMINARIES
In words, there exists an open interval I[a;b] centered at x0such that
f(x)>0 for allx2I:Hence, because f(x)0 we must have
Zb
af(x)dxZ
If(x)dx> 0
which contradicts our assumption that the integral is zero. We conclude that
f(x) = 0 on [a;b]:
(c) This follows from (b) by taking g(x) =f(x)
Remark 1.1
The above theorem remains valid for functions in two variables. For example,
iff(x;y) is de ned for xin an interval Iandyin an interval Jsuch that
Zb
aZd
cf(x;y)dxdy = 0
for all [a;b]Jand [c;d]Ithenf(x;y) = 0 over the rectangle IJ:
Example 1.1
Letf;g: [a;b]!Rbe continuous and such that f(x)g(x) for allxin
[a;b]:Show that ifRb
a(g(x) f(x))dx= 0 thenf(x)g(x) on [a;b]:
Solution.
Apply part (b) of previous theorem to the function h(x) =g(x) f(x)
Partial Derivatives
For multivariable functions, there are two common notations for partial
derivatives, and we shall employ them interchangeably. The rst is the Leib-
nitz notation that employs the symbol @to denote partial derivative. The
second, a more compact notation, is to use subscripts to indicate partial
derivatives. For example, utrepresents@u
@t;whileuxxrepresents@2u
@x2;anduxxt
becomes@3u
@2x@t:
An important formula of di erentiation is the so-called chain rule. If
u=u(x;y) wherex=x(s;t) andy=y(s;t) then
@u
@s=@u
@x@x
@s+@u
@y@y
@s:
1 SOME RESULTS OF CALCULUS 9
Likewise,
@u
@t=@u
@x@x
@t+@u
@y@y
@t:
Example 1.2
Compute the partial derivatives indicated:
(a)@
@y(y2sinxy)
(b)@2
@x2[ex+y]2
Solution.
(a) We have@
@y(y2sinxy) = sinxy@
@y(y2)+y2@
@y(sinxy) = 2ysinxy+xy2cosxy:
(b) We have@
@x[ex+y]2=@
@xe2(x+y)= 2e2(x+y):Thus,@2
@x2[ex+y]2=@
@x2e2(x+y)=
4e2(x+y)
Example 1.3
Supposeu(x;y) = sin (x2+y2);wherex=tesandy=s+t:Findusandut:
Solution.
We have
us=uxxs+uyys= 2xcos (x2+y2)tes+ 2ycos (x2+y2)
=[2tes+ 2(s+t)] cos [t2e2s+ (s+t)2]
Likewise,
ut=uxxt+uyyt= 2xcos (x2+y2)es+ 2ycos (x2+y2)
=[2tes+ 2(s+t)] cos [t2e2s+ (s+t)2]
Often we must di erentiate an integral with respect to a parameter which
may appear in the limits of integration, or in the integrand.
Letf(x;t) be a continuous function in the rectangle faxbgfct
dg:Assume that@f
@tis continuous on this rectangle. De ne the function
J(t) =Zb(t)
a(t)f(x;t)dx
wherea(t) andb(t) are continuously di erentiable functions of tsuch that
aa(t)b(t)b:Recall that a function f(x) is said to be continnu-
ously di erentiable if the derivative f0(x) exists, and is itself a continuous
function.
10 PRELIMINARIES
Theorem 1.2
dJ
dt=d
dtZb(t)
a(t)f(x;t)dx
=f(b(t);t)b0(t) f(a(t);t)a0(t) +Zb(t)
a(t)@f
@t(x;t)dx
Example 1.4
Consider the heat problem
ut=kuxx u; > 0; k> 0;0<x<L; t> 0
with boundary conditions ux(0;t) = 0 =ux(L;t) and initial condition u(x;0) =
f(x):LetE(t) =1
2RL
0u2dx:
(a) Show that E0(t)0:
(b) Show that E(t)RL
01
2jf(x)j2dx:
Solution.
(a) We have
dE
dt=1
2ZL
0@
@tu2(x;t)dx
=ZL
0u(x;t)ut(x;t)dx=kZL
0u(x;t)uxx(x;t)dx ZL
0u2(x;t)dx
=ku(x;t)ux(x;t)jL
0 kZL
0u2
x(x;t)dx ZL
0u2(x;t)dx
= kZL
0u2
x(x;t)dx ZL
0u2(x;t)dx0:
(b) From (a) we conclude that E(t) is a decreasing function of t>0:Thus,
E(t)E(0) =1
2ZL
0u2(x;0)dx=ZL
01
2jf(x)j2dx
The Least Upper Bound
A function f:D!Ris said to be bounded from above inDif there is a
constantMsuch thatf(x)Mfor allx2D:We callMan upper bound
1 SOME RESULTS OF CALCULUS 11
off:Note that the numbers, M+ 1;M+ 2;are also upper bounds of
f:The smallest upper bound of fis called the least upper bound or the
supremum. IfMis the supremum of finDwe write
M= supff(x) :x2Dg:
Note that if Nis any upper bound of finDthenMN:
Example 1.5
Find the supremum of f(x) = sinx:
Solution.
The graph of fis bounded between 1 and 1:Thus, supff(x) :x2Rg= 1
Example 1.6
Find
sup 2sinx
sint
:x2R; t> 0
Solution.
The answer is
sup 2sinx
sint
:x2R;t> 0
=2
12 PRELIMINARIES
Practice Problems
Exercise 1.1
Compute the partial derivatives indicated:
(a)@
@x(y2sinxy)
(b)@2
@x2(ex2y)
(c)@4
@x@y2@z
zln
x2
y
:
Exercise 1.2
Find all the rst partial derivatives of the functions:
(a)f(x;y) =x4+ 6py
(b)f(x;y;z ) =x2y 10y2z3+ 43x 7 tan (4y)
(c)f(s;t) =t7ln (s2) +9
t3 7p
s4
(d)f(x;y) = cos 4
x
ex2y 5y3
(e)f(u;v) =9u
u2+5v
(f)f(x;y;z ) =xsiny
z2
(g)f(x;y) =p
x2+ ln (5x 3y2)
Exercise 1.3
Letf(x;y) =e3xcosy:Computefx(0;2):
Exercise 1.4
Ifz=exsiny; x =st2;andy=s2t, nd@z
@sand@z
@t:
Exercise 1.5
In the equation
@u
@x @u
@y=x 2y
identify the independent variable(s) and the dependent variable.
Exercise 1.6
Letfbe an odd function, that is, f( x) = f(x) for allx2R:Show that
for alla2Rwe haveZa
af(x)dx= 0:
1 SOME RESULTS OF CALCULUS 13
Exercise 1.7
Letfbe an even function, that is, f( x) =f(x) for allx2R:Show that
for alla2Rwe have
Za
af(x)dx= 2Za
0f(x)dx:
Exercise 1.8
Use the product rule of derivatives to derive the formula of integration by
parts Z
uv0dx=uv Z
u0vdx:
Exercise 1.9
Letu(x;t) =2sin x
sin t
:Finduttanduxx:
Exercise 1.10
Letu(x;t) =2sin x
sinh t
;where
sinhx=ex e x
2:
Finduttanduxx:
Exercise 1.11
Find
sup 2sinht
sinx
:x2R
:
Exercise 1.12
Letun(x;t) = 1 +en2t
nsinnx.
(a) Find supfjun(x;0) 1j:x2Rg:
(b) Find supfjun(x;t) 1j:x2Rg:
14 PRELIMINARIES
2 Sequences of Functions: Pointwise and Uni-
form Convergence
Later in this book we will be constructing solutions to PDEs involving in nite
sums of sines and cosines. These in nite sums or series are called Fourier
series. Fourier series are examples of series of functions. Convergence of
series of functions is de ned in terms of convergence of a sequence of func-
tions. In this section we study the two types of convergence of sequences of
functions.
Recall that a sequence of numbers fang1
n=1is said to converge to a number
Lif and only if for every given >0 there is a positive integer N=N()
such that for all nNwe havejan Lj<:
What is the analogue concept of convergence when the terms of the sequence
are variables? Let DRand for each n2Nconsider a function fn:D!R:
Thus, we obtain a sequence of functions ffng1
n=1:For such a sequence, there
are two types of convergenve that we consider in this section: pointwise con-
vergence and uniform convergence.
We say thatffng1
n=1converges pointwise onDto a function f:D!Rif
and only if for a given a2Dand>0 there is a positive integer N=N(a;)
such that if nNthenjfn(a) f(a)j<:In symbol, we write
lim
n!1fn(a) =f(a):
It is important to note that Nis a function of both aand:
Example 2.1
De nefn: [0;1)!Rbyfn(x) =nx
1+n2x2:Show that the sequence ffng1
n=1
converges pointwise to the function f(x) = 0 for all x0:
Solution.
For allx0;
lim
n!1fn(x) = lim
n!1nx
1 +n2x2= 0
Example 2.2
For each positive integer nletfn: (0;1)!Rbe given by fn(x) =nx:Show
thatffng1
n=1does not converge pointwise on D= (0;1):
2 SEQUENCES OF FUNCTIONS: POINTWISE AND UNIFORM CONVERGENCE 15
Solution.
This follows from the fact that lim n!1nx=1for allx2D
As pointed out above, for pointwise convergence, the positive integer Nde-
pends on both the given xand:A stronger convergence concept can be
de ned where Ndepends only on :
LetDbe a subset of Rand letffng1
n=1be a sequence of functions de ned on
D:We say thatffng1
n=1converges uniformly onDto a function f:D!R
if and only if for all >0 there is a positive integer N=N() such that if
nNthenjfn(x) f(x)j<for allx2D:
This de nition says that the integer Ndepends only on the given so that
fornN, the graph of fn(x) is bounded above by the graph of f(x) +and
below by the graph of f(x) :
Example 2.3
For each positive integer nletfn: [0;1]!Rbe given by fn(x) =x
n:Show
thatffng1
n=1converges uniformly to the zero function.
Solution.
Let >0 be given. Let Nbe a positive integer such that N >1
:Then for
nNwe have
jfn(x) f(x)j=jxj
n1
n1
N<
for allx2[0;1]
Clearly, uniform convergence implies pointwise convergence to the same limit
function. However, the converse is not true in general.
Example 2.4
De nefn: [0;1)!Rbyfn(x) =nx
1+n2x2:By Example 2.1, this sequence
converges pointwise to f(x) = 0:Let=1
3:Show that there is no positive
integerNwith the property nNimpliesjfn(x) f(x)j<for allx0:
Hence, the given sequence does not converge uniformly to f(x):
Solution.
For any positive integer Nand fornNwe have
fn1
n
f1
n =1
2>
16 PRELIMINARIES
Exercise 2.1 below shows a sequence of continuous functions converging point-
wise to a discontinuous function. That is, pointwise convergence does not
preserve the property of continuity. One of the interesting features of uniform
convergence is that it preserves continuity as shown in the next example.
Example 2.5
Suppose that for each n1 the function fn:D!Ris continuous in D:
Suppose thatffng1
n=1converges uniformly to f:Leta2D:
(a) Let>0 be given. Show that there is a positive integer Nsuch that if
nNthenjfn(x) f(x)j<
3for allx2D:
(b) Show that there is a >0 such that for all jx aj<we havejfN(x)
fN(a)j<
3:
(c) Using (a) and (b) show that for jx aj< we havejf(x) f(a)j< :
Hence,fis continuous in Dsinceawas arbitrary. Symbolically we write
lim
x!alim
n!1fn(x) = lim
n!1lim
x!afn(x):
Solution.
(a) This follows from the de nition of uniform convergence.
(b) This follows from the fact that fNis continuous at a2D:
(c) Forjx aj<we havejf(x) f(a)j=jf(a) fN(a) +fN(a) fN(x) +
fN(x) f(x)jjfN(a) f(a)j+jfN(a) fN(x)j+jfN(x) f(x)j<
3+
3+
3=
Does pointwise convergenvce preserve integration? In real analysis, it is
proven that pointwise convergence does not preserve integrability. That is,
the pointwise limit of a sequence of integrable functions need not be inte-
grable. Even when a sequence of functions converges pointwise, the process
of interchanging limits and integration is not true in general.
Contrary to pointwise convergence, uniform convergence preserves integra-
tion. Moreover, limits and integration can be interchanged. That is, if
ffng1
n=1converges uniformly to fon a closed interval [ a;b] then
lim
n!1Zb
afn(x)dx=Zb
alim
n!1fn(x)dx:
Now, what about di erentiablility? Again, pointwise convergence fails in
general to conserve the di erentiability property. See Exercise 2.1. Does
uniform convergence preserve di erentiability? The answer is still no as
shown in the next example.
2 SEQUENCES OF FUNCTIONS: POINTWISE AND UNIFORM CONVERGENCE 17
Example 2.6
Consider the family of functions fn: [ 1;1] given by fn(x) =q
x2+1
n:
(a) Show that fnis di erentiable for each n1:
(b) Show that for all x2[ 1;1] we have
jfn(x) f(x)j1pn
wheref(x) =jxj:Hint: Note thatq
x2+1
n+p
x21pn:
(c) Let>0 be given. Show that there is a positive integer Nsuch that for
nNwe have
jfn(x) f(x)j<for allx2[ 1;1]:
Thus,ffng1
n=1converges uniformly to the non-di erentiable function f(x) =
jxj:
Solution.
(a)fnis the composition of two di erentiable functions so it is di erentiable
with derivative
f0
n(x) =x
x2+1
n 1
2
:
(b) We have
jfn(x) f(x)j= r
x2+1
n p
x2 = (q
x2+1
n p
x2)(q
x2+1
n+p
x2)
q
x2+1
n+p
x2
=1
nq
x2+1
n+p
x2
1
n
1pn=1pn
(c) Let >0 be given. Since lim n!11pn= 0 we can nd a positive integer
Nsuch that for all nNwe have1pn<: Now the answer to the question
follows from this and part (b)
Even when uniform convergence occurs, the process of interchanging lim-
its and di erentiation may fail as shown in the next example.
18 PRELIMINARIES
Example 2.7
Consider the functions fn:R!Rde ned byfn(x) =sinnx
n:
(a) Show thatffng1
n=1converges uniformly to the function f(x) = 0:
(b) Note thatffng1
n=1andfare di erentiable functions. Show that
lim
n!1f0
n(x)6=f0(x) =h
lim
n!1fn(x)i0
:
That is, one cannot, in general, interchange limits and derivatives.
Solution.
(a) Let>0 be given. Let Nbe a positive integer such that N >1
:Then
fornNwe have
jfn(x) f(x)j= sinnx
n 1
n<
and this is true for all x2R:Hence,ffng1
n=1converges uniformly to the
functionf(x) = 0:
(b) We have lim n!1f0
n() = limn!1cosn= limn!1( 1)nwhich does not
converge. However, f0() = 0
Pointwise convergence was not enough to preserve di erentiability, and nei-
ther was uniform convergence by itself. Even with uniform convergence the
process of interchanging limits with derivatives is not true in general. How-
ever, if we combine pointwise convergence with uniform convergence we can
indeed preserve di erentiability and also switch the limit process with the
process of di erentiation.
Theorem 2.3
Letffng1
n=1be a sequence of di erentiable functions on [ a;b] that converges
pointwise to some function fde ned on [ a;b]:Ifff0
ng1
n=1converges uniformly
on [a;b] to a function g;then the function fis di erentiable with derivative
equals tog:Thus,
lim
n!1f0
n(x) =g(x) =f0(x) =h
lim
n!1fn(x)i0
:
Finally, we conclude this section with the following important result that is
useful when a given sequence is bounded.
2 SEQUENCES OF FUNCTIONS: POINTWISE AND UNIFORM CONVERGENCE 19
Theorem 2.4
Consider a sequence fn:D!R:Then this sequence converges uniformly to
f:D!Rif and only if
lim
n!1supfjfn(x) f(x)j:x2Dg= 0:
Example 2.8
Show that the sequence de ned by fn(x) =cosx
nconverges uniformly to the
zero function.
Solution.
We have
0supfjcosx
nj:x2Rg1
n:
Now apply the squeeze rule for sequences we nd that
lim
n!1supfjcosx
nj:x2Rg= 0
which implies that the given sequence converges uniformly to the zero func-
tion on R
20 PRELIMINARIES
Practice Problems
Exercise 2.1
De nefn: [0;1]!Rbyfn(x) =xn:De nef: [0;1]!Rby
f(x) =0 if 0x<1
1 ifx= 1:
(a) Show that the sequence ffng1
n=1converges pointwise to f:
(b) Show that the sequence ffng1
n=1does not converge uniformly to f:Hint:
Suppose otherwise. Let = 0:5 and get a contradiction by using a point
(0:5)1
N<x< 1:
Exercise 2.2
Consider the sequence of functions
fn(x) =nx+x2
n2
de ned for all xinR:Show that this sequence converges pointwise to a
functionfto be determined.
Exercise 2.3
Consider the sequence of functions
fn(x) =sin (nx+ 3)pn+ 1
de ned for all xinR:Show that this sequence converges pointwise to a
functionfto be determined.
Exercise 2.4
Consider the sequence of functions de ned by fn(x) =n2xnfor all 0x1:
Show that this sequence does not converge pointwise to any function.
Exercise 2.5
Consider the sequence of functions de ned by fn(x) = (cosx)nfor all
2
x
2:Show that this sequence converges pointwise to a noncontinuous
function to be determined.
2 SEQUENCES OF FUNCTIONS: POINTWISE AND UNIFORM CONVERGENCE 21
Exercise 2.6
Consider the sequence of functions fn(x) =x xn
nde ned on [0 ;1):
(a) Doesffng1
n=1converge to some limit function? If so, nd the limit func-
tion and show whether the convergence is pointwise or uniform.
(b) Doesff0
ng1
n=1converge to some limit function? If so, nd the limit func-
tion and show whether the convergence is pointwise or uniform.
Exercise 2.7
Letfn(x) =xn
1+xnforx2[0;2]:
(a) Find the pointwise limit f(x) = limn!1fn(x) on [0;2]:
(b) Doesfn!funiformly on [0 ;2]?
Exercise 2.8
For eachn2Nde nefn:R!Rbyfn(x) =n+cosx
2n+sin2x:
(a) Show that fn!1
2uniformly.
(b) Find lim n!1R7
2fn(x)dx:
Exercise 2.9
Show that the sequence de ned by fn(x) = (cosx)ndoes not converge uni-
formly on [
2;
2]:
Exercise 2.10
Letffng1
n=1be a sequence of functions such that
supfjfn(x)j: 2x5g2n
1 + 4n:
(a) Show that this sequence converges uniformly to a function fto be found.
(b) What is the value of the limit lim n!1R5
2fn(x)dx?
22 PRELIMINARIES
Review of Some ODEs Results
Later on in this book, we will encounter problems where a given partial
di erential is reduced to an ordinary di erential function by means of a given
change of variables. Then techniques from the theory of ODE are required in
solving the transformed ODE. In this chapter, we include some of the results
from ODE theory that will be needed in our future discussions.
3 The Method of Integrating Factor
In this section, we discuss a technique for solving the rst order linear non-
homogeneous equation
y0+p(t)y=g(t) (3.1)
wherep(t) andg(t) are continuous on the open interval a<t<b:
Sincep(t) is continuous, it has an antiderivative namelyR
p(t)dt:Let(t) =
eR
p(t)dt:Multiply Equation (3.1) by (t) and notice that the left hand side of
the resulting equation is the derivative of a product. Indeed,
d
dt((t)y) =(t)g(t):
Integrate both sides of the last equation with respect to tto obtain
(t)y=Z
(t)g(t)dt+C
Hence,
y(t) =1
(t)Z
(t)g(t)dt+C
(t)
or
y(t) =e R
p(t)dtZ
eR
p(t)dtg(t)dt+Ce R
p(t)dt
23
24 REVIEW OF SOME ODES RESULTS
Notice that the second term of the previous expression is just the general
solution for the homogeneous equation
y0+p(t)y= 0
whereas the rst term is a solution to the nonhomogeneous equation. That
is, the general solution to Equation (3.1) is the sum of a particular solution of
the nonhomogeneous equation and the general solution of the homogeneous
equation.
Example 3.1
Solve the initial value problem
y0 y
t= 4t; y(1) = 5:
Solution.
We havep(t) = 1
tso that(t) =1
t:Multiplying the given equation by the
integrating factor and using the product rule we notice that
1
ty0
= 4:
Integrating with respect to tand then solving for ywe nd that the general
solution is given by
y(t) =tZ
4dt+Ct= 4t2+Ct:
Sincey(1) = 5;we ndC= 1 and hence the unique solution to the IVP is
y(t) = 4t2+t;0<t<1
Example 3.2
Find the general solution to the equation
y0+2
ty= lnt; t> 0:
Solution.
The integrating factor is (t) =eR2
tdt=t2:Multiplying the given equation
byt2to obtain
(t2y)0=t2lnt:
3 THE METHOD OF INTEGRATING FACTOR 25
Integrating with respect to twe nd
t2y=Z
t2lntdt+C:
The integral on the right-hand side is evaluated using integration by parts
withu= lnt;dv =t2dt;du =dt
t;v=t3
3obtaining
t2y=t3
3lnt t3
9+C
Thus,
y=t
3lnt t
9+C
t2
26 REVIEW OF SOME ODES RESULTS
Practice Problems
Exercise 3.1
Solve the IVP: y0+ 2ty=t; y(0) = 0:
Exercise 3.2
Find the general solution: y0+ 3y=t+e 2t:
Exercise 3.3
Find the general solution: y0+1
ty= 3 cost; t> 0:
Exercise 3.4
Find the general solution: y0+ 2y= cos (3t):
Exercise 3.5
Find the general solution: y0+ (cost)y= 3 cost:
Exercise 3.6
Given that the solution to the IVP ty0+ 4y= t2; y(1) = 1
3exists on the
interval 1<t<1:What is the value of the constant ?
Exercise 3.7
Suppose that y(t) =Ce 2t+t+ 1 is the general solution to the equation
y0+p(t)y=g(t):Determine the functions p(t) andg(t):
Exercise 3.8
Suppose that y(t) = 2e t+et+ sintis the unique solution to the IVP
y0+y=g(t); y(0) =y0:Determine the constant y0and the function g(t):
Exercise 3.9
Find the value (if any) of the unique solution to the IVP y0+ (1 + cost)y=
1 + cost; y(0) = 3 in the long run?
Exercise 3.10
Solve
aux+buy+cu= 0
by using the change of variables s=ax+byandt=bx ay:
3 THE METHOD OF INTEGRATING FACTOR 27
Sample Exam Questions
Exercise 3.11
Solve the initial value problem ty0=y+t; y(1) = 7:
Exercise 3.12
Show that if aandare positive constants, and bis any real number, then
every solution of the equation
y0+ay=be t
has the property that y!0 ast!1 . Hint: Consider the cases a=and
a6=separately.
Exercise 3.13
Solve the initial-value problem y0+y=ety2;y(0) = 1 using the substitution
u(t) =1
y(t)
Exercise 3.14
Solve the initial-value problem ty0+ 2y=t2 t+ 1; y(1) =1
2
Exercise 3.15
Solvey0 1
ty= sint; y (1) = 3:Express your answer in terms of the sine
integral ,Si(t) =Rt
0sins
sds:
28 REVIEW OF SOME ODES RESULTS
4 The Method of Separation of Variables for
ODEs
The method of separation of variables that you have seen in the theory of
ordinary di erential equations has an analogue in the theory of partial dif-
ferential equations (Section 17). In this section, we review the method for
ordinary di erentiable equations.
A rst order di erential equation is separable if it can be written with one
variable only on the left and the other variable only on the right:
f(y)y0=g(t)
To solve this equation, we proceed as follows. Let F(t) be an antiderivative
off(t) andG(t) be an antiderivative of g(t):Then by the Chain Rule
d
dtF(y) =dF
dydy
dt=f(y)y0
Thus,
f(y)y0 g(t) =d
dtF(y) d
dtG(t) =d
dt[F(y) G(t)] = 0
It follows that
F(y) G(t) =C
which is equivalent to
Z
f(y)y0dt=Z
g(t)dt+C
As you can see, the result is generally an implicit equation involving a func-
tion ofyand a function of t:It may or may not be possible to solve this to
getyexplicitly as a function of t:For an initial value problem, substitute the
values oftandybyt0andy0to get the value of C:
Remark 4.2
IfFis a di erentiable function of yandyis a di erentiable function of tand
bothFandyare given then the chain rule allows us to nddF
dtgiven by
dF
dt=dF
dydy
dt
For separable equations, we are given f(y)y0=dF
dtand we are asked to nd
F(y):This process is referred to as \reversing the chain rule."
4 THE METHOD OF SEPARATION OF VARIABLES FOR ODES 29
Example 4.1
Solve the initial value problem y0= 6ty2; y(1) =1
25:
Solution.
Separating the variables and integrating both sides we obtain
Zy0
y2dt=Z
6tdt
or
Zd
dt1
y
dt=Z
6tdt
Thus,
1
y(t)= 3t2+C
Sincey(1) =1
25;we ndC= 28:The unique solution to the IVP is then
given explicitly by
y(t) =1
28 3t2
Example 4.2
Solve the IVP yy0= 4 sin (2t); y(0) = 1:
Solution.
This is a separable di erential equation. Integrating both sides we nd
Zd
dty2
2
dt= 4Z
sin (2t)dt
Thus,
y2= 4 cos (2t) +C
Sincey(0) = 1;we ndC= 5:Now, solving explicitly for y(t) we nd
y(t) =p
4 cost+ 5
Sincey(0) = 1;we havey(t) =p 4 cost+ 5:The interval of existence of
the solution is the interval 1<t<1
30 REVIEW OF SOME ODES RESULTS
Practice Problems
Exercise 4.1
Solve the (separable) di erential equation
y0=tet2 lny2:
Exercise 4.2
Solve the (separable) di erential equation
y0=t2y 4y
t+ 2:
Exercise 4.3
Solve the (separable) di erential equation
ty0= 2(y 4):
Exercise 4.4
Solve the (separable) di erential equation
y0= 2y(2 y):
Exercise 4.5
Solve the IVP
y0=4 sin (2t)
y; y(0) = 1:
Exercise 4.6
Solve the IVP:
yy0= sint; y(
2) = 2:
Exercise 4.7
Solve the IVP:
y0+y+ 1 = 0; y(1) = 0:
Exercise 4.8
Solve the IVP:
y0 ty3= 0; y(0) = 2:
4 THE METHOD OF SEPARATION OF VARIABLES FOR ODES 31
Exercise 4.9
Solve the IVP:
y0= 1 +y2; y(
4) = 1:
Exercise 4.10
Solve the IVP:
y0=t ty2; y(0) =1
2:
32 REVIEW OF SOME ODES RESULTS
Sample Exam Questions
Exercise 4.11
For what values of the constants ;y 0;and integer nis the function y(t) =
(4 +t) 1
2a solution of the initial value problem?
y0+ yn= 0; y(0) =y0:
Exercise 4.12
Solve the equation 3 uy+uxy= 0 by using the substitution v=uy:
Exercise 4.13
Solve the IVP
(2y siny)y0= sint t; y(0) = 0:
Exercise 4.14
State an initial value problem, with initial condition imposed at t0= 2;
having implicit solution y3+t2+ siny= 4:
Exercise 4.15
Can the di erential equation
dy
dx=x2 xy
be solved by the method of separation of variables? Explain.
5 SECOND ORDER LINEAR ODES 33
5 Second Order Linear ODEs
When solving second order partial di erential equations such as the heat,
wave, and Laplace's equations using the method of separation of variables
for PDEs one ends up confronting second order linear ODEs. Thus, it is
deemed necessary to review some of the techniques used in solving second
order linear ordinary di erential equations which we do in this section.
We start rst by considering the second order linear ODE with constant
coecients given by
ay00+by0+cy= 0 (5.1)
wherea;bandcare constants with a6= 0:
Notice rst that for b= 0 andc6= 0 the function y00is a constant multiple
ofy:So it makes sense to look for a function with such property. One such
function is y(t) =ert:Substituting this function into (5.1) leads to
ay00+by0+cy=ar2ert+brert+cert= (ar2+br+c)ert= 0
Sinceert>0 for allt, the previous equation leads to
ar2+br+c= 0 (5.2)
Thus, a function y(t) =ertis a solution to (5.1) when rsatis es equation
(5.2). We call (5.2) the characteristic equation for (5.1) and the polyno-
mialC(r) =ar2+br+cis called the characteristic polynomial .
The characteristic equation is a quadratic equation. Thus, this equation can
have two distinct real solutions, two equal solutions, or two conjugate com-
plex solutions depending on the sign of the expression b2 4ac:Hence, we
consider the following three cases:
Case 1:b2 4ac> 0:
In this case, equation (5.2) have two distinct real roots r1= b p
b2 4ac
4aand
r2= b+p
b2 4ac
4a:The general solution to (5.1) is given by
y(t) =c1er1t+c2er2t
wherec1andc2are arbitrary constants.
Example 5.1
Solve the initial value problem
y00 y0 6y= 0; y(0) = 1; y0(0) = 2:
Describe the behavior of the solution y(t) ast! 1 andt!1:
34 REVIEW OF SOME ODES RESULTS
Solution.
The characteristic polynomial is C(r) =r2 r 6 = (r 3)(r+2) so that the
characteristic equation r2 r 6 = 0 has the solutions r1= 3 andr2= 2:
The general solution is then given by
y(t) =c1e3t+c2e 2t:
Taking the derivative to obtain
y0(t) = 3c1e3t 2c2e 2t:
The conditions y(0) = 1 and y0(0) = 2 lead to the system
c1+c2= 1
3c1 2c2= 2:
Solving this system by the method of elimination we nd c1=4
5andc2=1
5:
Hence, the unique solution to the initial value problem is
y(t) =1
5(4e3t+e 2t):
Ast! 1; e3t!0 ande 2t!1:Thus,y(t)!1:Similarly,y(t)!1
ast!1
Case 2:b2 4ac= 0:
In this case, the characteristic equation has the single root r= b
2a:The
general solution to (5.1) is given by
y(t) =c1e b
2at+c2te b
2at
wherec1andc2are arbitrary constants.
Example 5.2
Solve the initial value problem: y00+ 2y0+y= 0; y(0) = 1; y0
1(0) = 1:
Solution.
The characteristic equation r2+2r+1 = 0 has a repeated root: r1=r2= 1:
Thus, the general solution is given by
y(t) =c1e t+c2te t:
5 SECOND ORDER LINEAR ODES 35
The two conditions y(0) = 1 and y0(0) = 1 lead toc1= 1 andc2= 0:
Hence, the unique solution is y(t) =e t
Case 3:b2 4ac< 0:
In this case, the complex roots of equation (5.1) are given by
r1;2= bip
4ac b2
2a
wherei=p 1:The general solution is given by
y(t) =e t(c1cos t+c2sin t)
where = b
2a; =p
4ac b2
2a;andc1andc2are real numbers.
Example 5.3
Solve the initial value problem
y00 10y0+ 29y= 0; y(0) = 1; y0(0) = 3:
Solution.
The characteristic equation r2 10r+ 29 = 0 has the complex roots r1;2=
52i:Thus, the general solution is given by the expression
y(t) =e5t(c1cos 2t+c2sin 2t):
Findingy0we obtain
y0(t) =e5t[(5c1+ 2c2) cos 2t+ (5c2 2c1) sin 2t]:
The initial conditions yield c1= 1 andc2= 1:Thus, the unique solution
to the initial value problem is
y(t) =e5t(cos 2t sin 2t)
An Eigenvalue Problem
Consider the question of nding a nontrivial twice di erentiable function u
satisfying the ordinary di erential equation
d2u
dx2=u;0<x< 1:
36 REVIEW OF SOME ODES RESULTS
subject to the boundary conditions u(0) =u(1) = 0:This problem is referred
to as the eigenvalue problem for the following reason: De ne the function
Ld2
dx2:Then the given equation can be written as Lu=u:In linear
algebra,is called an eigenvalue ofLwith corresponding eigenvector u:
Di erent solutions to the eigenvalue problem are obtained depending on the
sign of:Suppose rst that = 0:Thenu(x) =C1x+C2for arbitrary
constantsC1andC2:Using the boundary conditions we nd C1=C2= 0:
Hence,u0:
Suppose that > 0:Thenu(x) =Aep
x+Be p
x:Again, the boundary
conditions imply that u0:
Now, suppose that < 0:Thenu(x) =Acosp
x+Bsinp
x:Using
the condition u(0) = 0 to obtain A= 0:Using the condition u(1) = 0
and assuming we are looking for non-trivial solution uwe expect to have
sinp
= 0:This happens when =n= (n)2wheren2N:We calln
an eigenvalue with corresponding eigenfunction un(x) = sinnx:
Finally, using the principle of superposition we nd that the general solution
to the eigenvalue problem is given by
u(x) =1X
n=1Ansinnx
where the convergence is pointwise convergence (See Section 2).
Euler Equations
A second order linear di erential equations of the form
ax2y00+bxy0+cy= 0
wherea;b;c are constants is called an Euler equation.
To solve Euler equation, one starts with solutions of the form y=xr(with
x > 0) whereris to be determined. Plugging this into the di erential
equation to get
ax2r(r 1)xr 2+bxrxr 1+cxr=0
(ar2 ar+br+c)xr=0
ar2 (a b)r+c=0
This last equation is a quadratic equation in rand so we will have three cases
to look at : Real distinct roots, double roots, and complex conjugate roots.
5 SECOND ORDER LINEAR ODES 37
If the quadratic equation has two distinct real roots r1andr2then the general
solution is given by
y(x) =Axr1+Bxr2:
If the quadratic equation has two equal roots r1=r2=rthen the general
solution is given by
y(x) =xr(A+Blnx):
If the quadratic equation has two complex conjugate solutions r1;2= i
then the general solution is given by
y(x) =x (Acos ( lnx) +Bsin ( lnx)):
Example 5.4
Solve the initial value problem
2x2y00+ 3xy0 15y= 0
y(1) = 0; y0(1) = 1:
Solution.
Lettingy=xrwe obtain the quadratic equation 2 r2+r 15 = 0 whose
roots arer1=5
2andr2= 3:Hence, the general solution is given by
y(x) =Ax5
2+Bx 3:
The condition y(1) = 0 implies A+B= 0:The condition y0(1) = 1 implies
5
2A 3B= 1:Solving this system of two unknowns we nd A=2
11and
B= 2
11:Hence, the unique solution is given by
y=2
11x5
2 2
11x 3
Second Order Linear nonhomogeneous ODE: The Method of Un-
determined Coecients
We consider the nonhomogeneous second order
ay00+by0+cy=g(t); a<t<b:
We know that the general solution has the structure
y(t) =c1y1(t) +c2y2(t) +yp(t)
38 REVIEW OF SOME ODES RESULTS
whereyp(t) is a particular solution to the nonhomogeneous equation. We
will writey(t) =yh(t) +yp(t) whereyh(t) =c1y1(t) +c2y2(t):
One way to nding ypis by using the method of undetermined coecients.
The idea behind the method of undetermined coecients is to look for yp(t)
which is of a form like that of g(t):This is possible only for special functions
g(t);but these special cases arise quite frequently in applications.
We will assume that g(t) being simple means it is some combination of terms
likeert;cos (kt);sin (kt);and polynomials antn+an 1tn 1+a1t+a0:Based
on those terms we will put together a candidate ypthat has some constants in
it we need to solve for: Those are the undetermined coecients this method
is named for.
In the following table we list examples of g(t) along with the corresponding
form of the particular solution.
Form ofg(t) Form ofyp(t)
antn+an 1tn 1++a1t+a0 tr[Antn+An 1tn 1++A1t+A0
[antn+an 1tn 1++a1t+a0]e ttr[Antn+An 1tn 1++A1t+A0]e t
[antn+an 1tn 1++a1t+a0] cos ttr[(Antn+An 1tn 1++A1t+A0) cos t
or +(Bntn+Bn 1tn 1++B1t+B0) sin t]
[antn+an 1tn 1++a1t+a0] sin t
e t[antn+an 1tn 1++a1t+a0] sin ttr[(Antn+An 1tn 1++A1t+A0)e tcos t
or +(Bntn+Bn 1tn 1++B1t+B0)e tsin t]
e t[antn+an 1tn 1++a1t+a0] cos t
The number ris chosen to be the smallest nonnegative integer such that
no term in the assumed form is a solution of the homogeneous equation
ay00+by0+cy= 0:The value of rwill be 0, 1, or 2.
Example 5.5
List an appropriate form for a particular solution of
(a)y00+ 4y=t2e3t:
(b)y00+ 4y=te2tcost:
(c)y00+ 4y= 2t2+ 5 sin 2t+e3t:
(d)y00+ 4y=t2cos 2t:
Solution.
The general solution to the homogeneous equation is yh(t) =c1cos 2t+
c2sin 2t:
(a) Forg(t) =t2e3t, an appropriate particular solution has the form yp(t) =
5 SECOND ORDER LINEAR ODES 39
tr(A2t2+A1t+A0)e3t:We taker= 0 since no term in the assumed form for
ypis present in the expression of yh(t):Thus
yp(t) = (A2t2+A1t+A0)e3t
(b) An appropriate form is
yp(t) =tr[(A1t+A0)e2tcost+ (B1t+B0)e2tsint]
We taker= 0 since no term in the assumed form for ypis present in the
expression of yh(t):Thus
yp(t) = (A1t+A0)e2tcost+ (B1t+B0)e2tsint
(c)
yp(t) =A2t2+A1t+A0+B0tcos 2t+C0tsin 2t+D0e3t
(d)
yp(t) =t(A2t2+A1t+A0) cos 2t+t(B2t2+B1t+B0) sin 2t
Example 5.6
Find the general solution of
y00 2y0 3y= 4t 5 + 6te2t
Solution.
The characteristic equation of the homogeneous equation is r2 2r 3 = 0
with rootsr1= 1 andr2= 3:Thus,
yh(t) =c1e t+c2e3t
A guess for the particular solution is yp(t) =At+B+Cte2t+De2t:Inserting
this into the di erential equation leads to
3At 2A 3B 3Cte2t+ (2C 3D)e2t= 4t 5 + 6te2t
From this identity we obtain 3A= 4 so that A= 4
3:Also, 2A 3B= 5
so thatB=23
9:Since 3C= 6 we nd C= 2:From 2C 3D= 0 we nd
D= 4
3:It follows that
y(t) =c1e t+c2e3t 4
3t+23
9
2t+4
3
e2t
40 REVIEW OF SOME ODES RESULTS
Practice Problems
Exercise 5.1
Solve the initial value problem
y00 4y0+ 3y= 0; y(0) = 1; y0(0) = 1
Describe the behavior of the solution y(t) ast! 1 andt!1:
Exercise 5.2
Solve the initial value problem
y00+ 4y0+ 2y= 0; y(0) = 0; y0(0) = 4
Describe the behavior of the solution y(t) ast! 1 andt!1:
Exercise 5.3
Solve the initial value problem
2y00 y= 0; y(0) = 2; y0(0) =p
2
Describe the behavior of the solution y(t) ast! 1 andt!1:
Exercise 5.4
Find a homogeneous second-order linear ordinary di erential equation whose
general solution is y(t) =c1e2t+c2e t:
Exercise 5.5
Solve the IVP
9y00 6y0+y= 0; y(3) = 2; y0(3) = 5
3
Exercise 5.6
Solve the IVP
25y00+ 20y0+ 4y= 0; y(5) = 4e 2; y0(5) = 3
5e 2
Exercise 5.7
The graph of a solution y(t) of the di erential equation 4 y00+ 4y0+y= 0
passes through the points (1 ;e 1
2) and (2;0):Determiney(0) andy0(0):
5 SECOND ORDER LINEAR ODES 41
Exercise 5.8
Find the general solution of y00 6y0+ 9y= 0:
Exercise 5.9
Solve the IVP
y00+ 2y0+ 2y= 0; y(0) = 3; y0(0) = 1
Exercise 5.10
Solve the IVP
2y00 2y0+y= 0; y( ) = 1; y0( ) = 1
Exercise 5.11
Find the general solution of
y00 y0+y= 2 sin 3t
Exercise 5.12
Find the general solution of
y00+ 4y0 2y= 2t2 3t+ 6
42 REVIEW OF SOME ODES RESULTS
Sample Exam Questions
Exercise 5.13
Find the general solution to the following di erential equation.
x2y00 7xy0+ 16y= 0:
Exercise 5.14
Find the general solution to the following di erential equation.
x2y00+ 3xy0+ 4y= 0:
Exercise 5.15
Consider the di erential equation
d2y
dx2+y= 0:
Determine the eigenvalues and the corresponding eigenfunctions if ysatis-
es the following boundary conditions:
(a)y(0) =y() = 0
(b)y(0) =y0(L) = 0
(c)y0(0) =y(1) = 0:
Exercise 5.16
Show by direct computation that the eigenvalue problems
(ky0(x))0+y(x) = 0; k> 0
with the following boundary conditions have no negative eigenvalues :
(a)y(0) =y(L) = 0
(b)y0(0) =y0(L) = 0
(c)y(L) =y( L); y0(L) =y0( L):
Exercise 5.17
Solve the initial-value problem: 2 y00+ 5y0 3y= 0; y(0) = 2; y0(0) = 1:
Exercise 5.18
Find the general solution of
y00 y0= 5et sin 2t
Exercise 5.19
Solve using undetermined coecients:
y00+y0 2y=t+ sin 2t;y(0) = 1;y0(0) = 0
Introduction to PDEs
Many elds in engineering and the physical sciences require the study of ODE
and PDE. Examples of those elds are acoustics, aerodynamics, elasticity,
electrodynamics,
uid dynamics, geophysics (seismic wave propagation), heat
transfer, meteorology, oceanography, optics, petroleum engineering, plasma
physics (ionized liquids and gases), quantum mechanics.
So the study of partial di erential equation is of great importance to the
above mentioned elds. The purpose of this chapter is to introduce the
reader to the basic terms of partial di erential equations.
6 The Basic Concepts
The goal of this section is to introduce the reader to the basic concepts and
notations that will be used in the remainder of this book.
Adi erential equation is an equation that involves an unknown scalar
function (the dependent variable) and one or more of its derivatives. For
example,
d2y
dx2 5dy
dx+ 3y= 3 (6.1)
or
@u
@t @2u
@x2 @2u
@y2+u= 0: (6.2)
If the unknown function is a function in one single variable then the di er-
ential equation is called an ordinary di erential equation. An example
of an ordinary di erential equation is Equation (6.1). In contrast, when the
unknown function is a function of two or more independent variables then
the di erential equation is called a partial di erential equation , in short
PDE. Equation (6.2) is an example of a partial di erential equation. In this
book we will be focusing on partial di erential equations.
43
44 INTRODUCTION TO PDES
Example 6.1
Identify which variables are dependent variable or independent variable(s)
for the following di erential equations.
(a)d4y
dx4 x2+y= 0
(b)utt+xutx= 0:
(c)xdx
dt= 4:
(d)@y
@u 4@y
@v=u+ 3y:
Solution.
(a) Independent variable is xand the dependent variable is y:
(b) Independent variables are xandtand the dependent variable is u:
(c) Independent variable is tand the dependent variable is x:
(d) Independent variables are uandvand the dependent variable is y
Example 6.2
Classify the following as either ODE or PDE.
(a)ut=c2uxx:
(b)y00 4y0+ 5y= 0:
(c)ut+cux= 5:
Solution.
(a) PDE (b) ODE (c) PDE
Theorder of a partial di erential equation is the highest order derivative
occurring in the equation. Thus, (6.2) is a second order partial di erential
equation.
Example 6.3
Find the order of each of the following partial di erential equations:
(a)xux+yuy=x2+y2
(b)uux+uy= 2
(c)utt c2uxx=f(x;t)
(d)ut+uux+uxxx= 0
(e)utt+uxxxx= 0:
Solution.
(a) First order (b) First order (c) Second order (d) Third order (e) Fourth
order
6 THE BASIC CONCEPTS 45
A partial di erential equation is called linear if it is linear in the unknown
function and all its derivatives with coecients depend only on the indepen-
dent variables. For example, a rst order linear partial di erential equation
has the form
A(x;y)ux+B(x;y)uy+C(x;y)u=D(x;y)
whereas a second order linear partial di erential equation has the form
A(x;y)uxx+B(x;y)uxy+C(x;y)uyy+D(x;y)ux+E(x;y)uy+F(x;y)u=G(x;y):
A partial di erential equation is called quasi-linear if the highest-order
derivatives which appear in the equation are of degree 1(regardless of the
manner in which lower-order derivatives and unknown functions occur in the
equation). For example, a rst order quasi-linear partial di erential equation
has the form
A(x;y;u )ux+B(x;y;u )uy=C(x;y;u )
whereas a second order quasi-linear partial di erential equation has the form
A(x;y;u;ux;uy)uxx+B(x;y;u;ux;uy)uxy+C(x;y;u;ux;uy)uyy=D(x;y;u;ux;uy):
A partial di erential equation is semi-linear if it is quasi-linear and the
coecients of the highest-order derivatives are functions of independent vari-
ables only. For example, a rst order semi-linear partial di erential equation
has the form
A(x;y)ux+B(x;y)uy=C(x;y;u )
whereas a second order semi-linear partial di erential equation has the form
A(x;y)uxx+B(x;y)uxy+C(x;y)uyy=D(x;y;u;u x;uy):
Note that linear and semi-linear partial di erential equations are special cases
of quasi-linear equations.
A partial di erential equation that is not linear is called nonlinear . For
example,u2
x+ 2uxy= 0:
As for ODEs, linear PDEs are usually simpler to analyze/solve than nonlinear
PDEs.
46 INTRODUCTION TO PDES
Example 6.4
Determine whether the given PDE is linear, quasilinear, semilinear, or non-
linear:
(a)xux+yuy=x2+y2
(b)uux+uy= 2
(c)utt c2uxx=f(x;t)
(d)ut+uux+uxxx= 0
(e)u2
tt+uxxxx= 0:
Solution.
(a) Linear, quasilinear, semilinear.
(b) Quasilinear, nonlinear.
(c) Linear, quasilinear, semilinear.
(d) Quasilinear, semilinear, nonlinear.
(e) Quasilinear, semilinear, nonlinear
A more precise de nition of a linear di erential equation begins with the
concept of a linear di erential operator L:The operator Lis assembled
by summing the basic partial derivative operators, with coecients depend-
ing on the independent variables. The operator acts on suciently smooth
functions depending on the relevant independent variables. Linearity im-
poses two key requirements:
L[u+v] =L[u] +L[v] andL[ u] = L[u];
for any two (suciently smooth) functions u; vand any constant :
Example 6.5
De ne a linear di erential operator for the PDE
ut=c2uxx:
Solution.
LetL[u] =ut c2uxx:Then one can easily check that L[u+v] =L[u] +L[v]
andL[ u] = L[u]
A linear partial di erential equation is called homogeneous if every term
of the equation involves the unknown function or its partial derivatives. A
linear partial di erential equation that is not homogeneous is called nonho-
mogeneous. In this case, there is a term in the equation that involves only
6 THE BASIC CONCEPTS 47
the independent variables.
A homogeneous linear partial di erential equation has the form
L[u] = 0
whereLis a linear di erential operator.
Example 6.6
Determine whether the equation is homogeneous or nonhomogeneous:
(a)xux+yuy=x2+y2:
(b)utt=c2uxx:
(c)uxx+uyy= 0:
Solution.
(a) Nonhomogeneous because of x2+y2:
(b) Homogeneous.
(c) Homogeneous
Finally, we shall be employing a few basic notational conventions regard-
ing the variables that appear in our di erential equations. We always use
tto denote time, while x;y;z will represent (Cartesian) space coordinates.
Polar coordinates r;will also be used when needed, and our notational con-
ventions appear at the appropriate places in the exposition.
Anequilibrium equation models an unchanging physical system, and so
only involves the space variables. The time variable tappears when mod-
eling dynamical , meaning time-varying, processes. Both time and space
coordinates are independent variables.
48 INTRODUCTION TO PDES
Practice Problems
Exercise 6.1
Classify the following equations as either ODE or PDE.
(a) (y000)4+t2
(y0)2+4= 0
(b)@u
@x+y@u
@y=y x
y+x
(c)y00 4y= 0
Exercise 6.2
Write the equation
uxx+ 2uxy+uyy= 0
in the coordinates s=x; t=x y:
Exercise 6.3
Write the equation
uxx 2uxy+ 5uyy= 0
in the coordinates s=x+y; t= 2x:
Exercise 6.4
For each of the following PDEs, state its order and whether it is linear or
nonlinear. If it is linear, also state whether it is homogeneous or nonhomo-
geneous:
(a)uux+x2uyyy+ sinx= 0
(b)ux+ex2uy= 0
(c)utt+ (siny)uyy etcosy= 0:
Exercise 6.5
For each of the following PDEs, determine its order and whether it is linear
or not. For linear PDEs, state also whether the equation is homogeneous or
not. For nonlinear PDEs, circle all term(s) that are not linear.
(a)x2uxx+exu=xuxyy
(b)eyuxxx+exu= siny+ 10xuy
(c)y2uxx+exuux= 2xuy+u
(d)uxuxxy+exuuy= 5x2ux
(e)ut=k2(uxx+uyy) +f(x;y;t ):
6 THE BASIC CONCEPTS 49
Exercise 6.6
Which of the following PDEs are linear?
(a)Laplace's equation: uxx+uyy= 0:
(b)Convection (transport) equation: ut+cux= 0:
(c)Minimal surface equation: (1+Z2
y)Zxx 2ZxZyZxy+(1+Z2
x)Zyy= 0:
(d)Korteweg-Vries equation: ut+ 6uux=uxxx:
Exercise 6.7
Classify the following di erential equations as ODEs or PDEs, linear or
nonlinear, and determine their order. For the linear equations, determine
whether or not they are homogeneous.
(a) The di usion equation foru(x;t) :
ut=kuxx:
(b) The wave equation forw(x;t) :
wtt=c2wxx:
(c) The thin lm equation forh(x;t) :
ht= (hhxxx)x:
(d) The forced harmonic oscillator fory(t) :
ytt+!2y=Fcos (!t):
(e) The Poisson Equation for the electric potential ( x;y;z ) :
xx+ yy+ zz= 4(x;y;z ):
where(x;y;z ) is a known charge density.
(f)Burger's equation forh(x;t) :
ht+hhx=hxx:
Exercise 6.8
Write down the general form of a linear second order di erential equation of
a function in three variables.
50 INTRODUCTION TO PDES
Exercise 6.9
Give the orders of the following PDEs, and classify them as linear or nonlin-
ear. If the PDE is linear, specify whether it is homogeneous or nonhomoge-
neous.
(a)x2uxxy+y2uyy log (1 +y2)u= 0
(b)ux+u3= 1
(c)uxxyy+exux=y
(d)uuxx+uyy u= 0
(e)uxx+ut= 3u:
Exercise 6.10
Consider the second-order PDE
uxx+ 4uxy+ 4uyy= 0:
Use the change of variables v(x;y) =y 2xandw(x;y) =xto show that
uww= 0:
6 THE BASIC CONCEPTS 51
Sample Exam Questions
Exercise 6.11
Write the one dimensional wave equation utt=c2uxxin the coordinates
v=x+ctandw=x ct:
Exercise 6.12
Write the PDE
uxx+ 2uxy 3uyy= 0
in the coordinates v(x;y) =y 3xandw(x;y) =x+y:
Exercise 6.13
Write the PDE
aux+buy= 0
in the coordinates s(x;y) =ax+byandt(x;y) =bx ay:Assumea2+b2>0:
Exercise 6.14
Write the PDE
ux+uy= 1
in the coordinates s=x+yandt=x y:
Exercise 6.15
Write the PDE
aut+bux=u; a;b6= 0
in the coordinates v=ax btandw=1
at:
52 INTRODUCTION TO PDES
7 Solutions and Related Topics
By aclassical solution orstrong solution to a partial di erential equation
we mean a function that satis es the equation. To solve a PDE is to nd all
its classical solutions. In the case of only two independent variables xandy;
a solutionu(x;y) is visualized geometrically as a surface, called a solution
surface or an integral surface in the (x;y;u ) space.
A formula that expresses all the solutions of a PDE is called the general
solution of the equation.
Example 7.1
Show that u(x;t) =e 2 2t(cosx sinx) is a solution to the equation
ut 2uxx= 0:
Solution.
Since
ut 2uxx= 2 2e 2 2t(cosx
sinx) 2e 2 2t( 2cosx+2sinx) = 0
the given function is a solution to the given equation
Example 7.2
Find the general solution of uxy= 0:
Solution.
Integrating rst we respect to ywe ndux(x;y) =f(x);wherefis an
arbitrary di erentiable function. Integrating uxwith respect to xwe nd
u(x;y) =R
f(x)dx+g(y);wheregis an arbitrary di erentiable function
Note that the general solution in the previous example involves two arbitrary
functions. In general, the general solution of a partial di erential equation
is an expression that involves arbitrary functions. This is in contrast to the
general solution of an ordinary di erential equation which involves arbitrary
constants.
Usually, a classical solution enjoys properties such as smootheness (i.e. a
function that has continuous derivatives up to some desired order over some
domain.) and continuity. However, in the theory of nonlinear pdes, there are
solutions that do not require the smoothness property. Such solutions are
7 SOLUTIONS AND RELATED TOPICS 53
called weak solutions orgeneralized solutions. We illustrate this con-
cept using equations rather than pdes. Consider the equation x2 y2= 0:
The function y=xis a classical solution of this equation. This solution is
in nitely di erentiable function. On the other hand, the function y=jxjis
also a solution to the given equation. However, this solution is not di eren-
tiable at 0. We call such a solution a weak solution. In this book, the word
solution will refer to a classical solution.
Example 7.3
Show thatu(x;t) =t+1
2x2is a classical solution to the PDE
ut=uxx: (7.1)
Solution.
Assume that the domain of de nition of uisDR2:Sinceu;ut;ux;utx;uxx
exist and are continuous in D(i.e.,uis smooth in D) andusatis es equation
(7.1), we conclude that uis a classical solution to the given PDE
Now, consider the linear di erential operator Las de ned in the previous
section. The de ning properties of linearity immediately imply the key facts
concerning homogeneous linear (di erential) equations.
Theorem 7.1
The sum of two solutions to a homogeneous linear di erential equation is
again a solution, as is the product of a solution by any constant.
Proof.
Letu1;u2be solutions, meaning that L[u1] = 0 andL[u2] = 0:Then, thanks
to linearity,
L[u1+u2] =L[u1] +L[u2] = 0;
and hence their sum u1+u2is a solution. Similarly, if is any constant, and
uany solution, then
L[ u] = L[u] = 0 = 0;
and so the scalar multiple uis also a solution
The following result is known as the superposition principle for homo-
geneous linear equations.
54 INTRODUCTION TO PDES
Theorem 7.2
Ifu1;;unare solutions to a common homogeneous linear partial di eren-
tial equation L[u] = 0;then the linear combination u=c1u1++cnunis
a solution for any choice of constants c1;;cn:
Proof.
The key fact is that, thanks to the linearity of L, for any suciently smooth
functionsu1;;unand any constants c1;;cn;
L[u] =L[c1u1++cnun] =L[c1u1++cn 1un 1] +L[cnun]
==L[c1u1] ++L[cnun] =c1L[u1] ++cnL[un]:
In particular, if the functions are solutions, so L[u1] = 0;;L[un] = 0;then
the right hand side of the above equation vanishes, proving that uis also a
solution to the homogeneous equation L[u] = 0
In physical applications, homogeneous linear equations model unforced sys-
tems that are subject to their own internal constraints. External forcing
is represented by an additional term that does not involve the dependent
variable. This results in the nonhomogeneous equation
L[u] =f
whereLis a linear partial di erential operator, uis the dependent variable,
andfis a given non-zero function of the independent variables alone.
You already learned the basic philosophy for solving of nonhomogeneous
linear equations in your study of elementary ordinary di erential equations.
Step one is to determine the general solution to the homogeneous equation.
Step two is to nd a particular solution to the nonhomogeneous version. The
general solution to the nonhomogeneous equation is then obtained by adding
the two together. Here is the general version of this procedure:
Theorem 7.3
Letuibe a particular solution to the nonhomogeneous linear equation L[u] =
f:Then the general solution to L[u] =fis given by u=ui+uh, whereuhis
the general solution to the corresponding homogeneous equation L[u] = 0:
Proof.
Let us rst show that u=ui+uhis also a solution to L[u] =f:By linearity,
L[u] =L[ui+uh] =L[ui] +L[uh] =f+ 0 =f:
7 SOLUTIONS AND RELATED TOPICS 55
To show that every solution to the nonhomogeneous equation can be ex-
pressed in this manner, suppose usatis esL[u] =f:Setuh=u ui:Then,
by linearity,
L[uh] =L[u ui] =L[u] L[ui] = 0;
and henceuhis a solution to the homogeneous di erential equation. Thus,
u=ui+uhhas the required form
In physical applications, one can interpret the particular solution uias a
response of the system to the external forcing function, while the solution
uhto the homogeneous equation represents the system's internal, unforced
motion. The general solution to a linear nonhomogeneous equation is thus a
combination of the external and internal responses.
As you have noticed by now, one solution of a linear PDE leads to the cre-
ation of lots of solutions. In contrast, nonlinear equations are much tougher
to deal with, for example, knowledge of several solutions does not necessarily
help in constructing others. Indeed, even nding one solution to a nonlinear
partial di erential equation can be quite a challenge.
In this introductory course, we will primarily but not exclusively con-
centrate on analyzing the most basic linear partial di erential equations. But
we will have occasion to brie
y foray into the nonlinear realm, to appreciate
some recent developments in this fascinating area of contemporary research
and applications.
As observed above, a general solution of a partial di erential equation has
in nitely many solutions. In almost all cases, this general solution is of little
use since it has to satisfy other supplementary conditions, usually called ini-
tial or boundary conditions. These conditions determine the unique solution
of interest.
Aboundary value problem is a partial di erential equation where either
the unknown function or its derivatives have values assigned on the physical
boundary of the domain in which the problem is speci ed. These conditions
are called boundary conditions . For example,
uxx+uyy=0 if 0 <x;y< 1
u(x;0) =u(x;1) =0 if 0 <x< 1
ux(0;y) =ux(1;y) =0 if 0 <y< 1:
56 INTRODUCTION TO PDES
There are three types of boundary conditions which arise frequently in for-
mulating physical problems:
1.Dirichlet Boundary Conditions: In this case, the dependent function
uis prescribed on the boundary of the bounded domain. For example, if the
bounded domain is the rectangular plate 0 < x < L 1and 0< y < L 2;the
boundary conditions u(0;y);u(L1;y);u(x;0);andu(x;L 2) are prescribed.
The boundary conditions are called homogeneous if the dependent variable
is zero at any point on the boundary, otherwise the boundary conditions are
called nonhomogeneous.
2.Neumann Boundary Conditions: In this case, rst partial derivatives
are prescribed on the boundary of the bounded domain. For example, the
Neuman boundary conditions for a rod of length L;where 0<x<L; are of
the formux(0;t) = andux(L;t) = ;where and are constants.
3.Robin or mixed Boundary Conditions: This occurs when the depen-
dent variable and its rst partial derivatives are prescribed on the boundary
of the bounded domain.
Aninitial valur problem (orCauchy problem ) is a partial di erential
equation together with a set of additional conditions on the solution or its
derivatives at either a given point or a given curve in the domain of the so-
lution. These conditions are called initial value conditions. For example,
thetransport equation
ut(x;t) +cux(x;t) =0
u(x;0) =f(x)
is a Cauchy problem.
It can be shown that initial conditions for a PDE are necessary and sucient
for the existence of a unique solution.
We say that an initial and/or boundary value problem associated with a PDE
iswell-posed if it has a solution which is unique and depends continuously
on the data given in the problem. The last condition, namely the continuous
dependence is important in physical problems. This condition means that
the solution changes by a small amount when the conditions change a little.
Such solutions are said to be stable .
Example 7.4
7 SOLUTIONS AND RELATED TOPICS 57
Forx2Randt>0 we consider the initial value problem
utt uxx=0
u(x;0) =ut(x;0) =0
Clearly,u(x;t) = 0 is a solution to this problem.
(a) Let 0<<< 1 be a very small number. Show that the function u(x;t) =
2sin x
sin t
is a solution to the problem
utt uxx=0
u(x;0) =0
ut(x;0) =sinx
(b) Show that sup fju(x;t) u(x;t)j:x2R;t > 0g=2:Thus, a small
change in the initial data leads to a small change in the solution. Hence, the
initial value problem is well-posed.
Solution.
(a) We have
@u
@t=sinx
cost
@2u
@t2= sinx
sint
@u
@x=cosx
sint
@2u
@x2= sinx
sint
Thus,@2u
@t2 @2u
@x2= 0:Moreover,u(x;0) = 0 and@
@tu(x;0) =sin x
:
(b) We have
supfju(x;t) u(x;t)j:x2R;t> 0g=2supf sinx
sint
:x2R;t> 0g
=2
A problem that is not well-posed is referred to as an ill-posed problem. We
illustrate this concept in the next example.
58 INTRODUCTION TO PDES
Example 7.5
Forx2Randt>0 we consider the initial value problem
utt+uxx=0
u(x;0) =ut(x;0) =0
Clearly,u(x;t) = 0 is a solution to this problem.
(a) Let 0<<< 1 be a very small number. Show that the function u(x;t) =
2sin x
sinh t
;where
sinhx=ex e x
2
is a solution to the problem
utt+uxx=0
u(x;0) =0
ut(x;0) =sinx
(b) Show that sup fj@
@tu(x;0) ut(x;0)j:x2Rg=and supfju(x;t)
u(x;t)j:x2Rg=2 sinh t
:
(c) Find lim t!1supfju(x;t) u(x;t)j:x2Rg:
Solution.
(a) We have
@u
@t=sinx
cosht
@2u
@t2= sinx
sinht
@u
@x=cosx
sinht
@2u
@x2= sinx
sinht
Thus,@2u
@t2+@2u
@x2= 0:Moreover,u(x;0) = 0 and@
@tu(x;0) =sin x
:
(b) We have
supfj@
@tu(x;0) ut(x;0)j:x2Rg= supf sinx
:x2Rg
=supf sinx
:x2Rg=
7 SOLUTIONS AND RELATED TOPICS 59
and
supfju(x;t) u(x;t)j:x2Rg=2supf sinht
sinx
:x2Rg
=2 sinht
:
(c) We have
lim
t!1supfju(x;t) u(x;t)j:x2Rg= lim
t!12 sinht
=1:
Thus, a small change in the initial data leads to a catastrophically change in
the solution. Hence, the given problem is ill-posed
60 INTRODUCTION TO PDES
Practice Problems
Exercise 7.1
Determineaandbso thatu(x;y) =eax+byis a solution to the equation
uxxxx+uyyyy+ 2uxxyy= 0:
Exercise 7.2
Consider the following di erential equation
tuxx ut= 0:
Supposeu(t;x) =X(x)T(t):Show that there is a constant such that
X00=XandT00=tT:
Exercise 7.3
Consider the initial value problem
xux+ (x+ 1)yuy= 0; x;y> 1
u(1;1) =e:
Show thatu(x;y) =xex
yis the solution to this problem.
Exercise 7.4
Show thatu(x;y) =e 2ysin (x y) is the solution to the initial value prob-
lem ux+uy+ 2u= 0 forx;y> 0
u(x;0) = sinx
Exercise 7.5
Solve each of the following di erential equations:
(a)du
dx= 0 whereu=u(x):
(b)@u
@x= 0 whereu=u(x;y):
Exercise 7.6
Solve each of the following di erential equations:
(a)d2u
dx2= 0 whereu=u(x):
(b)@2u
@x@y= 0 whereu=u(x;y):
7 SOLUTIONS AND RELATED TOPICS 61
Exercise 7.7
Show thatu(x;y) =f(y+ 2x) +xg(y+ 2x);wherefandgare two arbitrary
twice di erentiable functions, satisfy the equation
uxx 4uxy+ 4uyy= 0:
Exercise 7.8
Find the di erential equation whose general solution is given by u(x;t) =
f(x ct)+g(x+ct);wherefandgare arbitrary twice di erentiable functions
in one variable.
Exercise 7.9
Letp:R!Rbe a di erentiable function in one variable. Prove that
ut=p(u)ux
has a solution satisfying u(x;t) =f(x+p(u)t);wherefis an arbitrary
di erentiable function. Then nd the general solution to ut= (sinu)ux:
Exercise 7.10
Find the general solution to the pde
uxx+ 2uxy+uyy= 0:
Hint: See Exercise 6.2.
62 INTRODUCTION TO PDES
Sample Exam Questions
Exercise 7.11
Letu(x;t) be a function such that uxxexists andu(0;t) =u(L;t) = 0 for all
t2R:Prove thatZL
0uxx(x;t)u(x;t)dx0:
Exercise 7.12
Consider the initial value problem
ut+uxx= 0; x2R; t> 0
u(x;0) = 1:
(a) Show that u(x;t)1 is a solution to this problem.
(b) Show that un(x;t) = 1 +en2t
nsinnxis a solution to the initial value
problem
ut+uxx= 0; x2R; t> 0
u(x;0) = 1 +sinnx
n:
(c) Find supfjun(x;0) 1j:x2Rg:
(d) Find supfjun(x;t) 1j:x2Rg:
(e) Show that the problem is ill-posed.
Exercise 7.13
Find the general solution of each of the following PDEs by means of direct
integration.
(a)ux= 3x2+y2; u=u(x;y):
(b)uxy=x2y; u =u(x;y):
(c)uxyz= 0; u=u(x;y;z ):
(d)uxtt=e2x+3t; u=u(x;t):
Exercise 7.14
Consider the second-order PDE
uxx+ 4uxy+ 4uyy= 0:
(a) Use the change of variables v(x;y) =y 2xandw(x;y) =xto show
thatuww= 0:
(b) Find the general solution to the given PDE.
7 SOLUTIONS AND RELATED TOPICS 63
Exercise 7.15
Derive the general solution to the PDE
utt=c2uxx
by using the change of variables v=x+ctandw=x ct:
64 INTRODUCTION TO PDES
First Order Partial Di erential
Equations
Many problems in the mathematical, physical, and engineering sciences deal
with the formulation and the solution of rst order partial di erential equa-
tions. Our rst task is to understand simple rst order equations. In ap-
plications, rst order partial di erential equations are most commonly used
to describe dynamical processes, and so time, t;is one of the independent
variables. Most of our discussion will focus on dynamical models in a single
space dimension, bearing in mind that most of the methods can be readily
extended to higher dimensional situations. First order partial di erential
equations and systems model a wide variety of wave phenomena, including
transport of solvents in
uids,
ood waves, acoustics, gas dynamics, glacier
motion, trac
ow, and also a variety of biological and ecological systems.
From a mathematical point of view, rst order partial di erential equations
have the advantage of providing conceptual basis that can be utilized in the
study of higher order partial di erential equations.
In this chapter we introduce the basic de nitions of rst order partial dif-
ferential equations. We then derive the one dimensional spatial transport
eqution and discuss some methods of solutions. One general method of solv-
ability for quasilinear rst order partial di erential equation, known as the
method of characteristics, is analyzed.
8 Classi cation of First Order PDEs
In this section, we present the basic de nitions pertained to rst order PDE.
By a rst order di erential equation in two variables xandywe mean
65
66 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
any equation of the form
F(x;y;u;u x;uy) = 0: (8.1)
In what follows the functions a;b; andcare assumed to be continuously
di erentiable functions. If Equation (8.1) can be written in the form
a(x;y;u )ux+b(x;y;u )uy=c(x;y;u ) (8.2)
then we say that the equation is quasilinear . The following are examples
of quasilinear equations:
uux+uy+cu2= 0
x(y2+u)ux y(x2+u)uy= (x2 y2)u:
If Equation (8.1) can be written in the form
a(x;y)ux+b(x;y)uy=c(x;y;u ) (8.3)
then we say that the equation is semilinear . The following are examples of
semilinear equations:
xux+yuy=u2+x2
(x+ 1)2ux+ (y 1)2uy= (x+y)u2:
If Equation (8.1) can be written in the form
a(x;y)ux+b(x;y)uy+c(x;y)u=d(x;y) (8.4)
then we say that the equation is linear . Examples of linear equations are:
xux+yuy=cu
(y z)yx+ (z x)uy+ (x y)uz= 0:
A rst order pde that is not linear is said to be nonlinear. Examples of
nonlinear equations are:
ux+cu2
y=xy
u2
x+u2
y=c:
First order partial di erential equations are classi ed as either linear or non-
linear. Clearly, linear equations are a special kind of quasilinear equation
8 CLASSIFICATION OF FIRST ORDER PDES 67
(8.2) ifaandbare functions of xandyonly andcis a linear function of u:
Likewise, semilinear equations are quasilinear equations if aandbare func-
tions ofxandyonly. Also, semilinear equations (8.4) reduces to a linear
equation if cis linear in u:
A linear equation is called homogeneous ifd(x;y)0 and nonhomoge-
neous ifd(x;y)6= 0:Examples of linear homogeneous equations are:
xux+yuy=cu
(y z)ux+ (z x)uy+ (x y)uz= 0:
Examples of nonhomogeneous equations are:
ux+ (x+y)uy u=ex
yux+xuy=xy:
Recall that for an ordinary linear di erential equation, the general solution
depends mainly on arbitrary constants. Unlike ODEs, in linear partial dif-
ferential equations, the general solution depends on arbitrary functions.
Example 8.1
Solve the equation ut(x;t) = 0:
Solution.
The general solution is given by u(x;t) =f(x) wherefis an arbitrary dif-
ferentiable function of x
Example 8.2
Consider the transport equation
aut(x;t) +bux(x;t) = 0
whereaandbare constants. Show that u(x;t) =f(bt ax) is a solution
to the given equation, where fis an arbitrary di erentiable function in one
variable.
Solution.
Letv(x;t) =bt ax:Using the chain rule we see that ut(x;t) =bfv(v) and
ux(x;t) = afv(v):Hence,aut(x;t) +bux(x;t) =abfv(v) abfv(v) = 0
68 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
Practice Problems
Exercise 8.1
Classify each of the following PDE as linear, quasilinear, semi-linear, or non-
linear.
(a)xux+yuy= sin (xy):
(b)ut+uux= 0
(c)u2
x+u3u4
y= 0:
(d) (x+ 3)ux+xy2uy=u3:
Exercise 8.2
Show thatu(x;y) =exf(2x y);wherefis a di erentiable function of one
variable, is a solution to the equation
ux+ 2uy u= 0:
Exercise 8.3
Show thatu(x;y) =xpxysatis es the equation
xux yuy=u
subject to
u(y;y) =y2; y0:
Exercise 8.4
Show thatu(x;y) = cos (x2+y2) satis es the equation
yux+xuy= 0
subject to
u(0;y) = cosy2:
Exercise 8.5
Show thatu(x;y) =y 1
2(x2 y2) satis es the equation
1
xux+1
yuy=1
y
subject tou(x;1) =1
2(3 x2):
8 CLASSIFICATION OF FIRST ORDER PDES 69
Exercise 8.6
Find a relationship between aandbifu(x;y) =f(ax+by) is a solution to
the equation 3 ux 7uy= 0 for any di erentiable function f:
Exercise 8.7
SupposeLis a linear operator, that is, L( u+ v) = L(u)+ L(v):Consider
the homogeneous and nonhomogeneous linear equations
Lu= 0
Lu=f
wherefis some function. Suppose vis a solution to the homogeneous equa-
tion, andwis a solution to the nonhomogeneous equation. Show u=av+w
is a solution to the nonhomogeneous equation for any constant a:
Exercise 8.8
Reduce the partial di erential equation
aux+buy+cu= 0
to a rst order ODE by introducing the change of variables s=ax+byand
t=bx ay:
Exercise 8.9
Solve the partial di erential equation
ux+uy= 1
by introducing the change of variables s=x+yandt=x y:
70 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
Sample Exam Questions
Exercise 8.10
Show thatu(x;y) =e 4xf(2x 3y) is a solution to the rst-order PDE
3ux+ 2uy+ 12u= 0:
Exercise 8.11
Derive the general solution of the PDE
aut+bux=u; a;b6= 0
by using the change of variables v=ax btandw=1
at:
Exercise 8.12
Derive the general solution of the PDE
aux+buy= 0; a;b6= 0
by using the change of variables s(x;y) =ax+byandt(x;y) =bx ay:
Assumea2+b2>0:
Exercise 8.13
Write the equation
ut+cux+u=f(x;y)
in the coordinates v=x ct; w =t:
Exercise 8.14
Suppose that u(x;t) =w(x ct) is a solution to the PDE
xux+tut=Au
whereAandcare constants. Let v=x ct:Write the di erential equation
with unknown function w(v):
9 THE ONE DIMENSIONAL SPATIAL TRANSPORT EQUATIONS 71
9 The One Dimensional Spatial Transport Equa-
tions
Modeling is the process of writing a di erential equation to describe a physi-
cal situation. In this section we discuss the one-dimensional transport equa-
tion and discuss an analytical method for solving it.
Linear Transport Equation for Fluid Flows
We shall describe the transport of a dissolved chemical by water that is trav-
eling with uniform velocity cthrough a long thin tube Gwith uniform cross
sectionA:(The very same discussion applies to the description of the trans-
port of gas by air moving through a pipe.) We identify Gwith the open
interval (a;b);and the velocity c>0 is in the (rightward) positive direction
of thex axis. We will assume that the concentration of the chemical is con-
stant across the cross section Aat each point xso that the chemical changes
in thex direction and thus the term one-dimensional spatial equation. See
Figure 9.1
Figure 9.1
Letu(x;t) be a continuously di erentiable function denoting the concentra-
tion of the chemical (i.e. amount of chemical/area) at position xat time
t:Then at time t;the amount of chemical stored in a section of the tube
between positions aandxis given by the de nite integral
Zx
aAu(s;t)ds:
72 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
Since the water is
owing at a speed c;so at timeh+tthe same quantity of
chemical will be
Zx
aAu(s;t)ds=Zx+ch
a+chAu(s;t+h)ds:
Taking the derivative of both sides with respect to xwe nd
u(x;t) =u(x+ch;t+h):
Now taking the derivative of this last equation with respect to hwe nd
0 =ut(x+ch;t+h) +cux(x+ch;t+h):
Taking the limit of this last equation as happroaches 0 we nd
ut(x;t) +cux(x;t) = 0 (9.1)
for all (x;t):This equation is called the transport equation in one-dimensional
space. It is a linear, homogeneous rst order partial di erential equation.
Example 9.1
Show thatu(x;t) =f(x ct) is a solution to (9.1), where fis an arbitrary
di erentiable function in one variable.
Solution.
Using the chain rule we nd
ut= cf0(x ct) andux=f0(x ct):
Hence, by substituting these results into the equation we nd
ut+cux= cf0(x ct) +cf0(x ct) = 0:
The solution u(x;t) =f(x ct) is called the right traveling wave, since
the graph of the function f(x ct) at a given time tis the graph of f(x)
shifted to the right by the value ct:Thus, with growing time, the function
f(x) is moving without changes to the right at the speed c
An initial value condition determines a unique solution to the transport equa-
tion as stated in the next theorem.
9 THE ONE DIMENSIONAL SPATIAL TRANSPORT EQUATIONS 73
Theorem 9.4
Letgbe a continuously di erentiable function. Then there is a unique con-
tinuously di erentiable solution u(x;t) to the IVP
aux(x;t) +but(x;t) = 0
u(x;0) =g(x):
Indeed,uis given explicitly by the formula
u(x;t) =f(bx at); g(x) =f(bx):
Method of Solutions: The Coordinate Method
We will solve (9.1) by solving the more general equation
aux+buy= 0 (9.2)
wherea2+b2>0:
We introduce a new rectangular system by the substitution
s=ax+by; t =bx ay
According to the chain rule for the derivative of a composite function, we
have
ux=ussx+uttx=aus+but
uy=ussy+utty=bus aut
Substituting these into (9.2) to obtain
a2us+abut+b2us abut= 0
or
(a2+b2)us= 0
and sincea2+b2>0 we obtain
us= 0:
Solving this equation, we nd
u(s;t) =f(t)
wherefis an arbitrary di erentiable function of one variable. Now, in terms
ofxandywe nd
u(x;y) =f(bx ay):
74 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
Example 9.2
Use the coordinate method to nd the solution to ut 3ux= 0; u(x;0) =e x2:
Solution.
Letv= 3x+tandw=x+ 3t:Thenux= 3uv+uwandut=uv+ 3uw:
Substituting these into the given equation we nd 10 uv= 0 oruv= 0:
Hence,u(v;w) =f(w) oru(x;t) =f(x+ 3t) wherefis a di erentiable
function in one variable. Since u(x;0) =e x2;we nde x2=f(x):Hence,
u(x;t) =e (x+3t)2
Transport Equation with Decay: The Method of Characteristic
Coordinates
Atransport equation with decay is an equation given by
ut+cux+u=f(x;t)
whereandcare constants and fis a given function representing external
resources. Note that the decay is characterized by the term u:
To solve this equation, we introduce the characteristic coordinates given
by
v=x ct; w =t:
Using the chain rule, we nd
ut=uvvt+uwwt= cuv+uw
ux=uvvx+uwwx=uv:
Substituting these into the original equation we obtain the equation
uw+u=f(v+cw;w )
which can be solved by the method of integrating factor. We illustrate this
approach in the next example.
Example 9.3
Find the general solution of the transport equation
ut+ux u=t:
9 THE ONE DIMENSIONAL SPATIAL TRANSPORT EQUATIONS 75
Solution.
The characteristic coordinates are
v=x t; w =t:
These transform the original equation to the rst order ODE
uw u=w:
Using the method of integrating factor, we nd
d
dw(e wu) =we w
and solving this equation we nd
u(v;w) = (1 +w) +ewf(v)
and in terms of xandtwe nd
u(x;t) =f(x t)et (1 +t)
A more general method for solving quasilinear rst order partial di erential
equations, known as the method of characteristics, will be discussed in the
next section.
76 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
Practice Problems
Exercise 9.1
Use the coordinate method to nd the solution to ut+ 3ux= 0; u(x;0) =
sinx:
Exercise 9.2
Use the coordinate method, solve the equation aux+buy+cu= 0:
Exercise 9.3
Use the coordinate method, solve the equation ux+ 2uy= cos (y 2x) with
the initial condition u(0;y) =f(y);wheref:R!Ris a given function.
Exercise 9.4
Show that the initial value problem ut+ux=x; u(x;x) = 1 has no solution.
Exercise 9.5
Solve the transport equation ut+ 2ux= 3uwith initial condition u(x;0) =
1
1+x2:
Exercise 9.6
Solveut+ux 3u=twith initial condition u(x;0) =x2:
Exercise 9.7
Show that the decay term uin the transport equation with decay
ut+cux+u= 0
can be eliminated by the substitution w=uet:
Exercise 9.8
Use the coordinate method to solve
ux+uy=u2
u(x;0) =h(x)
9 THE ONE DIMENSIONAL SPATIAL TRANSPORT EQUATIONS 77
Exercise 9.9 (Well-Posed)
Letube the unique solution to the IVP
ut+cux= 0
u(x;0) =f(x)
andvbe the unique solution to the IVP
ut+cux= 0
u(x;0) =g(x)
wherefandgare continuously di erentiable functions.
(a) Show that w(x;t) =u(x;t) v(x;t) is the unique solution to the IVP
ut+cux= 0
u(x;0) =f(x) g(x)
(b) Write an explicit formula for win terms of fandg:
(c) Use (b) to conclude that the transport problem is well-posed. That is, a
small change in the initial data leads to a small change in the solution.
Exercise 9.10
Solve the initial boundary value problem
ut+cux= u; x> 0; t> 0
u(x;0) = 0; u(0;t) =g(t); t> 0:
78 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
Sample Exam Questions
Exercise 9.11
Solve the rst-order equation 2 ut+3ux= 0 with the initial condition u(x;0) =
sinx:
Exercise 9.12
Solve the PDE
ux+uy= 1
using the coordinate method.
Exercise 9.13
Consider the rst order linear homogeneous PDE
Aux+Buy+Cu= 0
whereA;B; andCare constants with A6= 0:
(a) Determine a;b;c;d in terms of A;B;C such thatad bc6= 0 and so that
the change of variables v=ax+byandw=cx+dywill reduce the given
PDE to a rst order PDE of the form uv+ u= 0:
(b) Use (a) to nd the general solution of the given PDE.
Exercise 9.14
Use the result of the previous problem to solve the PDE
ux+uy+u= 0:
10 THE METHOD OF CHARACTERISTICS 79
10 The Method of Characteristics
In this section we develop a method for nding the general solution of a
quasilinear rst order partial di erential equation. This method is called
themethod of characteristics orLagrange's method. This method of
solution can be described by the following result.
Theorem 10.1
The general solution of the quasilinear rst order PDE
a(x;y;u )ux+b(x;y;u )uy=c(x;y;u ) (10.1)
is given by
f(v;w) = 0 (10.2)
wherefis an arbitrary di erentiable function of v(x;y;u ) andw(x;y;u ) and
v=constant= c1; w=constant= c2are solutions to the ODE system
dx
a=dy
b=du
c: (10.3)
Equations (10.3) are called the characteristic equations in non-parametric
forms. The corresponding parametric forms are given by the system of ODEs
dx
ds=a
dy
ds=b
du
ds=c
Remark 10.1
Sometimes (10.2) is written explicitly as v=g(w) orw=g(v) wheregis an
arbitrary di erentiable function.
Example 10.1
Find the general solution of the PDE x2ux+y2uy= (x+y)u:
Solution.
The characteristic equations for this PDE aredx
x2=dy
y2=du
(x+y)u:Using the
80 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
rst two fractions, we haveRdx
x2=Rdy
y2and this impliesx y
xy=c1:Also,
we can solve for xobtainingx=1
1
y c1andx+y=y
1 c1y+y:Using the
last two fractions we nddy
y2=du
u(x+y)=)1
y2
y
1 c1y+y
dy=du
u=)
1
y c1y2+1
y
dy=du
u=)
1 c1y
y c1y2+c1
1 c1y+1
y
dy=du
u=)R
2
y+c1
1 c1y
dy=
Rdu
u=)2 lnjyj lnj1 c1yj= lnjuj+c2=)y2
1 c1y=cu=)yy
1 c1y=
cu=)y1
1
y c1=cu=)xy=cu:Hence, the general solution is
fx y
xy;xy
u
= 0
wherefis an arbitrary di erentiable function
Example 10.2
Find the general solution of the PDE yuux+xuuy=xy:
Solution.
The characteristic equations aredx
yu=dy
xu=du
xy:Using the rst two fractions
we ndx2 y2=c1:Using the last two fractions we nd u2 y2=c2:Hence,
the general solution is f(x2 y2;u2 y2) = 0 oru2=y2+g(x2 y2);where
fandgare arbitrary di erentiable functions
Example 10.3
Find the general solution of the PDE x(y2 u2)ux y(u2+x2)yy= (x2+y2)u:
Solution.
The characteristic equations aredx
x(y2 u2)=dy
y(u2+x2)=du
(x2+y2)u:Using a
property of proportions we can write
xdx+ydy+udu
x2(y2 u2) y2(u2+x2) +u2(x2+y2)=du
(x2+y2)u:
That is
xdx+ydy+udu
0=du
(x2+y2)u
or
xdx+ydy+udu= 0:
10 THE METHOD OF CHARACTERISTICS 81
Hence, we nd x2+y2+u2=c1:Also,
dx
x dy
y
y2 u2+u2+x2=du
(x2+y2)u
or
dx
x dy
y=du
u:
Hence, we nd ln yu
x =constant oryu
x=c2:The general solution is given by
f
x2+y2+u2;yu
x
= 0
or
u=x
yg(x2+y2+u2)
wherefandgare arbitrary di erentiable functions
Example 10.4
Solve the transport equation using the method of characteristics
ut+cux= 0:
Solution.
The characteristic equations are given by
dt
1=dx
c=du
0:
Solving the rst two fractions we nd x ct=k:The last fraction implies
u=k0:The general solution is given by f(x ct;u) = 0 oru=g(x ct)
Solution curves to the ODE
dy
dx=b
a
are called characteristic curves or simply characteristics. These are
curves in the xy plane.
Example 10.5
Find the characteristics of cos yux+uy+xu= 0:
82 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
Solution.
Solving the equationdy
dx=1
cosyby the separation of variable method we nd
siny x=k
Example 10.6
Find the characteristics of ux+ 2uy u= 0:
Solution.
We havea= 1 andb= 2:Thus,dy
dx= 2 so that the characteristics are given
by 2x y=k
10 THE METHOD OF CHARACTERISTICS 83
Practice Problem
Exercise 10.1
Find the characteristics of the PDE
xux yuy=u:
Exercise 10.2
Find the characteristics of the PDE
yux+xuy= 0:
Exercise 10.3
Find the characteristics of the PDE
(x+y)(ux+uy) =u 1:
Exercise 10.4
Find the general solution of the PDE xux+yuy= 1 +u2:
Exercise 10.5
Find the general solution of the PDE ln ( y+u)ux+uy= 1:
Exercise 10.6
Find the general solution of the PDE xux+yuy=u:
Exercise 10.7
Find the general solution of the PDE xux+yuy=nu:
Exercise 10.8
Find the general solution of the PDE x(y u)ux+y(u x)uy=u(x y):
Exercise 10.9
Find the general solution of the PDE u(u2+xy)(xux yuy) =x4:
Exercise 10.10
Find the general solution of the PDE ( y+xu)ux (x+yu)uy=x2 y2:
Exercise 10.11
Find the general solution of the PDE ( y2+u2)ux xyuy+xu= 0:
84 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
Exercise 10.12
Find the general form of solutions to
ux+ 2uy=u
and sketch some of the characteristics. Hint: de ne a new variable v=e xu:
What equation does vsatisfy?
Exercise 10.13
Find the general form of solutions to
(1 +x2)ux+uy= 0
and sketch some of the characteristics.
10 THE METHOD OF CHARACTERISTICS 85
Sample Exam Questions
Exercise 10.14
Find the general solution of the equation
ux+yuy=u:
Exercise 10.15
Find the characteristics associated with the PDE
ux+xuy+ 3u= 2:
Exercise 10.16
Find the general solution of the rst order PDE
ux+yuy+xu= 0:
Exercise 10.17
Find the characteristics of the PDE
1
xux+1
yuy= 0:
Exercise 10.18
Find the characteristics of the PDE
1
xux+1
yuy=1
y:
86 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
11 The Cauchy Problem for First Order Quasi-
linear Equations
When solving a partial di erential equation, it is seldom the case that one
tries to study the properties of the general solution of such equations. In
general, one deals with those partial di erential equations whose solutions
satisfy certain supplementary conditions. In the case of a rst order partial
di erential equation, we determine the particular solution by formulating an
initial value porblem also known as a Cauchy problem.
In this section, we discuss the Cauchy problem for the rst order quasilinear
partial di erential equation
a(x;y;u )ux+b(x;y;u )uy=c(x;y;u ): (11.1)
Recall that the initial value problem of a rst order ordinary di erential
equation asks for a solution of the equation which has a given value at a
given point in R:The Cauchy problem for the PDE (11.1) asks for a solution
of (11.1) which has given values on a given curve in R2:A precise statement
of the problem is given next.
Initial Value Problem or Cauchy Problem
LetCbe a given curve in R2de ned parametrically by the equations
x=x0(t); y=y0(t)
wherex0;y0are continuously di erentiable functions on some interval I:Let
u0(t) be a given continuously di erentiable function on I:The Cauchy prob-
lem for (11.1) asks for a continuously di erentiable function u=u(x;y)
de ned in a domain
R2containing the curve Cand such that:
(1)u=u(x;y) is a solution of (11.1) in
.
(2) On the curve C; u equals the given function u0(t);i.e.
u(x0(t);y0(t)) =u0(t); t2I: (11.2)
We callCtheinitial curve of the problem, u0(t) the initial data , and
(11.2) the initial condition of the problem. See Figure 11.1.
11 THE CAUCHY PROBLEM FOR FIRST ORDER QUASILINEAR EQUATIONS 87
Figure 11.1
If we view a solution u=u(x;y) of (11.1) as an integral surface of (11.1),
we can give a simple geometrical statement of the problem: Find a solu-
tion surface of (11.1) containing the curve described parametrically by the
equations
:x=x0(t); y=y0(t); u=u0(t); t2I:
Note that the projection of this curve in the xy plane is just the curve C:
The following theorem asserts that under certain conditions the Cauchy prob-
lem (11.1) - (11.2) has a unique solution.
Theorem 11.1
Suppose that x0(t);y0(t);andu0(t) are continuously di erentiable functions
oftin an interval I;and thata;b;andcare functions of x;y; anduwith
continuous rst order partial derivatives with respect to their argument in
some domain Dof (x;y;u ) space containing the initial curve
:x=x0(t); y=y0(t); u=u0(t)
wheret2I:Then for each point ( x0(t);y0(t);u0(t)) on that satis es the
condition
a(x0(t);y0(t);u0(t))dy0
dt(t) b(x0(t);y0(t);u0(t))dx0
dt(t)6= 0: (11.3)
there exists a unique solution u=u(x;y) of (11.1) in a neighborhood Uof
(x0(t);y0(t)) such that the initial condition (11.2) is satis ed for every point
onCcontained in U:See Figure 11.2.
88 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
Figure 11.2
Note that condition (11.3) implies that
dy0(t)
dx0(t)6=b(x0;y0;u0)
a(x0;y0;u0)
which means that the vector ( a(x0;y0;u0);b(x0;y0;u0);c(x0;y0;u0)) is not
tangent to :(Recall that the normal vector to Chas components
dy0(t)
dt; dx0(t)
t
so that a vector ( a;b) is tangent to Cifady0(t)
dt bdx0(t)
dt= 0:) It follows that
the Cauchy problem has a unique solution if Cis nowhere characteristic.
We construct the desired solution using the method of characteristics as fol-
lows: Pick a point ( x0(t);y0(t);u0(t))2 :Using this as the initial value
we solve the system of ODEs consisting of the characteristic equations in
parametric form
dx
ds=a(x(s);y(s);u(s))
dy
ds=b(x(s);y(s);u(s))
du
ds=c(x(s);y(s);u(s))
satisfying the initial condition
(x(0);y(0);u(0)) = (x0(t);y0(t);u0(t)):
11 THE CAUCHY PROBLEM FOR FIRST ORDER QUASILINEAR EQUATIONS 89
The solution depends on the parameter sso it consists of a triples of functions
x=x(s;t); y=y(s;t); u=u(s;t): (11.4)
This system represents the parametric representation of the integral surface
of the problem in which the curve corresponds to s= 0:The solution uis
recovered by solving the rst two equations in (11.4) for
t=t(x;y); s=s(x;y)
and substituting these into the third equation to obtain u(x;y) =u(s(x;y);t(x;y)):
Example 11.1
Solve the Cauchy problem
ux+uy=1
u(x;0) =f(x):
Solution.
The initial curve in R3can be given parametrically as
:x0(t) =t; y 0(t) = 0; u0(t) =f(t):
We have
a(x0(t);y0(t);u0(t))dy0
dt(t) b(x0(t);y0(t);u0(t))dx0
dt(t) = 16= 0
so by the above theorem the given Cauchy problem has a unique solution.
To nd this solution, we solve the system of ODEs
dx
ds=1
dy
ds=1
du
ds=1:
Solving this system we nd
x(s;t) =s+ (t); y(s;t) =s+ (t); u(s) =s+
(t):
90 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
Butx(0;t) =tso that (t) =t:Similarly,y(0;t) = 0 so that (t) = 0
andu(0;t) =f(t) implies
(t) =f(t):Hence, the unique solution is given
parametrically by the equations
x(s;t) =t+s; y (s;t) =s; u (s;t) =s+f(t):
Solving the rst two equations for sandtwe nd
s=y; t =x y
and substituting these into the third equation we nd
u(x;y) =y+f(x y):
Alternative Computation
We can apply the results of the previous section to nd the unique solution.
If we solve the characteristic equations in non-parametric form
dx
1=dy
1=du
1
we ndx y=c1andu x=c2:Thus, the general solution of the PDE
is given by u=x+F(x y):Using the Cauchy data u(x;0) =f(x) we nd
f(x) =x+F(x) which implies that F(x) =f(x) x:Hence, the unique
solution is given by
u(x;y) =x+f(x y) (x y) =y+f(x y)
If condition (11.3) is not satis ed than Cis a characteristic curve. If the
curve satis es the characteristic equations than the problem has in nitely
many solutions. To see this, pick an arbitrary point P0= (x0;y0;u0) on :
Pick a new initial curve 0passing through P0which is not tangent to at
P0:In this case, condition (11.3) is satis ed and the new Cauchy problem has
a unique solution. Since there are in nitely many ways of selecting 0;we
obtain in nitely many solutions. We illustrate this case in the next example.
Example 11.2
Solve the Cauchy problem
ux+uy=1
u(x;x) =x:
11 THE CAUCHY PROBLEM FOR FIRST ORDER QUASILINEAR EQUATIONS 91
Solution.
The initial curve in R3can be given parametrically as
:x0(t) =t; y 0(t) =t; u 0(t) =t:
We have
a(x0(t);y0(t);u0(t))dy0
dt(t) b(x0(t);y0(t);u0(t))dx0
dt(t) = 0:
As in Example 11.1, the general solution of the PDE is u(x;y) =y+f(x
y) wherefis an arbitrary di erentiable function. Using the Cauchy data
u(x;x) =xwe ndf(0) = 0:Thus, the solution is given by
u(x;y) =y+f(x y)
wherefis an arbitrary function such that f(0) = 0:There are in nitely
many choices for f:Hence, the problem has in nitely many solutions. Note
that satis es the characteristic equations
If condition (11.3) is not satis ed and if does not satisfy the characteristic
equations then it can be shown that the Cauchy problem has no solutions.
We illustrate this case next.
Example 11.3
Solve the Cauchy problem
ux+uy=1
u(x;x) =1:
Solution.
The initial curve in R3can be given parametrically as
:x0(t) =t; y 0(t) =t; u 0(t) = 1:
We have
a(x0(t);y0(t);u0(t))dy0
dt(t) b(x0(t);y0(t);u0(t))dx0
dt(t) = 0:
Solving the characteristic equations in parametric form we nd
x(s;t) =s+ (t); y(s;t) =s+ (t); u(s;t) =s+
(t):
92 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
Clearly, does not satisfy the characteristic equations. Now, the general
solution to the PDE is given by u=y+f(x y):Using the Cauchy data
u(x;x) = 1 we nd f(0) = 1 x;which is not possible since the LHS is a
xed number whereas the RHS is a variable expression. Hence, the problem
has no solutions
Example 11.4
Solve the Cauchy problem
ux uy=1
u(x;0) =x2: (11.5)
Solution.
The initial curve is given parametrically by
:x0(t) =t; y 0(t) = 0; u 0(t) =t2:
We have
a(x0(t);y0(t);u0(t))dy0
dt(t) b(x0(t);y0(t);u0(t))dx0
dt(t) = 16= 0
so the Cauchy problem has a unique solution.
The characteristic equations are
dx
1=dy
1=du
1:
Using the rst two fractions we nd x+y=c1:Using the rst and the third
fractions we nd u x=c2:Thus, the general solution can be represented
by
u=x+f(x+y)
wherefis an arbitrary di erentiable function. Using the Cauchy data
u(x;0) =x2we ndx2 x=f(x):Hence, the unique solution is given
by
u=x+ (x+y)2 (x+y) = (x+y)2 y
Example 11.5
Solve the initial value problem
ut+uux=x; u (x;0) = 1:
11 THE CAUCHY PROBLEM FOR FIRST ORDER QUASILINEAR EQUATIONS 93
Solution.
The initial curve is given parametrically by
:x0(t) =t; y 0(t) = 0; u 0(t) = 1:
We have
a(x0(t);y0(t);u0(t))dy0
dt(t) b(x0(t);y0(t);u0(t))dx0
dt(t) = 16= 0
so the Cauchy problem has a unique solution.
The characteristic equations are
dt
1=dx
u=du
x:
Since
dt
1=d(x+u)
x+u
we nd that ( x+u)e t=c1:Now, using the last two fractions we nd
u2 x2=c2:Hence, the general solution is given by
f((x+u)e t;u2 x2) = 0
wherefis an arbitrary di erentiable function. Using the Cauchy data we
ndc1= 1 +xandc2= 1 x2= 2(1 +x) (1 +x)2= 2c1 c2
1:Thus,
u2 x2= 2(x+u)e t (x+u)2e 2t
or
u x= 2e t (x+u)e 2t:
This can be reduced further as follows: u+ue 2t=x+ 2e t xe 2t=
2e t+x(1 e 2t) =)u=2e t
1+e 2t+x1 e 2t
1+e 2t= sech(t) +xtanh(t)
Example 11.6
Solve the initial value problem
uux+uy= 1
with the initial curve
:x0(t) = 2t2; y 0(t) = 2t; u 0(t) = 0; t> 0:
94 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
Solution.
We have
a(x0(t);y0(t);u0(t))dy0
dt(t) b(x0(t);y0(t);u0(t))dx0
dt(t) = 4t6= 0; t> 0
so the Cauchy problem has a unique solution.
The characteristic equations in parametric form are given by the system of
ODEs
dx
ds=u
dy
ds=1
du
ds=1:
Thus, the solution of this system depends on two parameters sandt:Solving
the last two equations we nd
y(s;t) =s+ (t); u(s;t) =s+
(t):
Solving the rst equation with ubeing replaced by s+
(t) we nd
x(s;t) =1
2s2+
(t)s+ (t):
Using the initial conditions
x(0;t) = 2t2; y(0;t) = 2t; u(0;t) = 0
we nd
x(s;t) =1
2s2+ 2t2; y(s;t) =s+ 2t; u(s;t) =s:
Eliminating sandtwe nd
(u y)2+u2= 2x:
Solving this quadratic equation in uto nd
2u=y(4x y2)1
2:
The solution surface satisfying u= 0 ony2= 2xis given by
2u=y (4x y2)1
2:
This represents a solution surface only when y2<4x:The solution does not
exist fory2>4x
11 THE CAUCHY PROBLEM FOR FIRST ORDER QUASILINEAR EQUATIONS 95
Practice Problems
Exercise 11.1
Solve
(y u)ux+ (u x)uy=x y
with the condition u(x;1
x) = 0:
Exercise 11.2
Solve the linear equation
yux+xuy=u
with the Cauchy data u(x;0) =x3:
Exercise 11.3
Solve
x(y2+u)ux y(x2+u)uy= (x2 y2)u
with the Cauchy data u(x; x) = 1:
Exercise 11.4
Solve
xux+yuy=xe u
with the Cauchy data u(x;x2) = 0:
Exercise 11.5
Solve the initial value problem
xux+uy= 0; u(x;0) =f(x)
using the characteristic equations in parametric form.
Exercise 11.6
Solve the initial value problem
ut+aux= 0; u(x;0) =f(x):
Exercise 11.7
Solve the initial value problem
aux+uy=u2; u(x;0) = cosx
96 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
Exercise 11.8
Solve the initial value problem
ux+xuy=u; u (1;y) =h(y):
Exercise 11.9
Solve the initial value problem
uux+uy= 0; u(x;0) =f(x):
Exercise 11.10
Solve the initial value problem
p
1 x2ux+uy= 0; u(0;y) =y:
11 THE CAUCHY PROBLEM FOR FIRST ORDER QUASILINEAR EQUATIONS 97
Sample Exam Questions
Exercise 11.11
Consider
xux+ 2yuy= 0:
(i) Find and sketch the characteristics.
(ii) Find the solution with u(1;y) =ey:
(iii) What happens if you try to nd the solution satisfying either u(0;y) =
g(y) oru(x;0) =h(x) for given functions gandh?
(iv) Explain, using your picture of the characteristics, what goes wrong at
(x;y) = (0;0):
Exercise 11.12
Solve the equation ux+uy=usubject to the condition u(x;0) = cosx:
Exercise 11.13
(a) Find the general solution of the equation
ux+yuy=u:
(b) Find the solution satisfying the Cauchy data u(x;3ex) = 2:
(c) Find the solution satisfying the Cauchy data u(x;ex) =ex:
Exercise 11.14
Solve the Cauchy problem
ux+ 4uy=x(u+ 1)
u(x;5x) = 1:
Exercise 11.15
Solve the Cauchy problem
ux uy=u
u(x; x) = sinx:
98 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS
Exercise 11.16
(a) Find the characteristics of the equation
yux+xuy= 0:
(b) Sketch some of the characteristics.
(c) Find the solution satisfying the boundary condition u(0;y) =e y2:
(d) In which region of the plane is the solution uniquely determined?
Exercise 11.17
Consider the equation ux+yuy= 0:Is there a solution satisfying the extra
condition
(a)u(x;0) = 1
(b)u(x;0) =x?
If yes, give a formula; if no, explain why.
Second Order Linear Partial
Di erential Equations
In this chapter we consider the three fundamental second order linear partial
di erential equations of parabolic, hyperbolic, and elliptic type. These types
arise in many applications such as the wave equation, the heat equation
and the Laplace's equation. We will study the solvability of each of these
equations.
12 Second Order PDEs in Two Variables
In this section we will brie
y review second order partial di erential equa-
tions.
Asecond order partial di erential equation in the variables xandy
is an equation of the form
F(x;y;u;u x;uy;uxx;uyy;uxy) = 0: (12.1)
If Equation (12.1) can be written in the form
A(x;y;u;ux;uy)uxx+B(x;y;u;ux;uy)uxy+C(x;y;u;ux;uy)uyy=D(x;y;u;ux;uy)
(12.2)
then we say that the equation is quasilinear .
If Equation (12.1) can be written in the form
A(x;y)uxx+B(x;y)uxy+C(x;y)uyy=D(x;y;u;u x;uy) (12.3)
then we say that the equation is semilinear .
If Equation (12.1) can be written in the form
A(x;y)uxx+B(x;y)uxy+C(x;y)uyy+D(x;y)ux+E(x;y)uy+F(x;y)u=G(x;y)
(12.4)
99
100SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
then we say that the equation is linear .
A linear equation is said to be homogeneous whenG(x;y)0 and non-
homogeneous otherwise.
Equation (12.4) resembles the general equation of a conic section
Ax2+Bxy +Cy2+Dx+Ey+F= 0
which is classi ed as either parabolic, hyperbolic, or elliptic based on the sign
of the discriminant B2 4AC: We do the same for a second order linear
partial di erential equation:
Hyperbolic: This occurs if B2 4AC > 0 at a given point in the domain
ofu:
Parabolic: This occurs if B2 4AC= 0 at a given point in the domain
ofu:
Elliptic: This occurs if B2 4AC < 0 at a given point in the domain of
u:
Example 12.1
Determine whether the equation uxx+xuyy= 0 is hyperbolic, parabolic or
elliptic.
Solution.
Here we are given A= 1;B= 0;andC=x:SinceB2 4AC= 4x;the
given equation is hyperbolic if x<0, parabolic if x= 0 and elliptic if x>0
Second order partial di erential equations arise in many areas of scienti c
applications. In what follows we list some of the well-known models that are
of great interest:
1. The heat equation in one-dimensional space is given by
ut=kuxx
wherekis a constant.
2. The wave equation in one-dimensional space is given by
utt=c2uxx
wherecis a constant.
3. The Laplace equation is given by
u=uxx+uyy= 0:
12 SECOND ORDER PDES IN TWO VARIABLES 101
Practice Problems
Exercise 12.1
Classify each of the following equation as hyperbolic, parabolic, or elliptic:
(a) Wave propagation: utt=c2uxx; c> 0:
(b) Heat conduction: ut=cuxx; c> 0:
(c) Laplace's equation: u=uxx+uyy= 0:
Exercise 12.2
Classify the following linear scalar PDE with constant coecents as hyper-
bolic, parabolic or elliptic.
(a)uxx+ 4uxy+ 5uyy+ux+ 2uy= 0:
(b)uxx 4uxy+ 4uyy+ 3ux+ 4u= 0:
(c)uxx+ 2uxy 3uyy+ 2ux+ 6uy= 0:
Exercise 12.3
Find the region(s) in the xy plane where the equation
(1 +x)uxx+ 2xyuxy y2uyy= 0
is elliptic, hyperbolic, or parabolic. Sketch these regions.
Exercise 12.4
Show thatu(x;t) = cosxsintis a solution to the problem
utt=uxx
u(x;0) = 0
ut(x;0) = cos x
ux(0;t) = 0
for allx;t> 0:
Exercise 12.5
Classify each of the following PDE as linear, quasilinear, semi-linear, or non-
linear.
(a)ut+uux=uuxx
(b)xutt+tuxx+u3u2
x=t+ 1
(c)utt=c2uxx
(d)u2
tt+ux= 0:
102SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Exercise 12.6
Show that, for all ( x;y)6= (0;0); u(x;y) = ln (x2+y2) is a solution of
uxx+uyy= 0;
and that, for all ( x;y;z )6= (0;0;0); u(x;y;z ) =1p
x2+y2+z2is a solution of
uxx+uyy+uzz= 0:
Exercise 12.7
Consider the eigenvalue problem
uxx=u; 0<x<L
ux(0) =k0u(0)
ux(L) = kLu(L)
with Robin boundary conditions, where k0andkLare given positive numbers
andu=u(x):Can this system have a nontrivial solution u60 for>0?
Hint: Multiply the rst equation by uand integrate over x2[0;L]:
Exercise 12.8
Show that u(x;y) =f(x)g(y), wherefandgare arbitrary di erentiable
functions, is a solution to the PDE
uuxy=uxuy:
Exercise 12.9
Show that for any n2N;the function un(x;y) = sinnxsinhnyis a solution
to the Laplace equation
u=uxx+uyy= 0:
Exercise 12.10
Solve
uxy=xy:
12 SECOND ORDER PDES IN TWO VARIABLES 103
Sample Exam Questions
Exercise 12.11
Classify each of the following second-oder PDEs according to whether they
are hyperbolic, parabolic, or elliptic:
(a) 2uxx 4uxy+ 7uyy u= 0:
(b)uxx 2 cosxuxy sin2xuyy= 0:
(c)yuxx+ 2(x 1)uxy (y+ 2)uyy= 0:
Exercise 12.12
Letc>0:By computing ux;uxx;ut;anduttshow that
u(x;t) =1
2(f(x+ct) +f(x ct)) +1
2cZx+ct
x ctg(s)ds
is a solution to the PDE
utt=c2uxx
wherefis twice di erentiable function and gis a di erentiable function.
Then compute and simplify u(x;0) andut(x;0):
Exercise 12.13
Consider the second-order PDE
yuxx+uxy x2uyy ux u= 0:
Determine the region DinR2;if such a region exists, that makes this PDE:
(a) hyperbolic, (b) parabolic, (c) elliptic.
Exercise 12.14
Consider the second-order hyperbolic PDE
uxx+ 2uxy 3uyy= 0:
Use the change of variables v(x;y) =y 3xandw(x;y) =x+yto solve the
given equation.
Exercise 12.15
Solve the Cauchy problem
uxx+ 2uxy 3uyy= 0:
u(x;2x) = 1; ux(x;2x) =x:
104SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
13 Hyperbolic Type: The Wave equation
The wave equation has many physical applications from sound waves in air
to magnetic waves in the Sun's atmosphere. However, the simplest systems
to visualize and describe are waves on a stretched elastic string.
Initially the string is horizontal with two xed ends say a left end Land a
right endR:Then from end Lwe shake the string and we notice a wave
propogate through the string. The aim is to try and determine the vertical
displacement from the x axis of the string, u(x;t);as a function of position
xand timet:A displacement of a tiny piece of the string between points P
andQis shown in Figure 13.1.
Figure 13.1
where
(x;t) is the angle between the string and a horizontal line at position x
and timet;
T(x;t) is the tension in the string at position xand timet;
(x) is the mass density of the string at position x:
To derive the wave equation we need to make some simplifying assumptions:
(1) The density of the string, ;is constant so that the mass of the string
betweenPandQis simplytimes the length of the string between Pand
13 HYPERBOLIC TYPE: THE WAVE EQUATION 105
Q, where the length of the string is sgiven by
s=p
(x)2+ (u)2= xs
1 +u
x2
xs
1 +@u
@x2
(2) The displacement, u(x;t), and its derivatives are assumed small so that
sx
and the mass of the portion of the string is
x:
(3) The only forces acting on this portion of the string are the tensions
T(x;t) atPandT(x+ x;t) atQ:(In physics, tension is the magnitude of
the pulling force exerted by a string). The gravitational force is neglected.
(4) Our tiny string element moves only vertically. Then the net horizontal
force on it must be zero.
Next, we consider the forces acting on the typical string portion shown in
Figure 13.1. These forces are:
(i) tension pulling to the right, which has magnitude T(x+ x;t);and acts
at an angle (x+ x;t) above the horizontal.
(ii) tension pulling to the left, which has magnitude T(x;t);and acts at an
angle(x;t) above the horizontal.
Now we resolve the forces into their horizontal and vertical components.
Horizontal: The net horizontal force of the tiny string is T(x+x;t) cos(x+ x;t)
T(x;t) cos(x;t):Since there is no horizontal motion, we must have
T(x;t) cos(x;t) =T(x+ x;t) cos(x+ x;t) =T: (13.1)
Vertical: AtPthe tension force is T(x;t) sin(x;t) whereas at Qthe
force isT(x+ x;t) sin(x+ x;t). Then Newton's Law of motion
massacceleration = Applied Forces
gives
x@2u
@t2=T(x+ x;t) sin(x+ x;t) T(x;t) sin(x;t):
106SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Dividing by Tand using (13.1) we obtain
Tx@2u
@t2=T(x+ x;t) sin(x+ x;t)
T(x+ x;t) cos(x+ x;t) T(x;t) sin(x;t)
T(x;t) cos(x;t)
= tan(x+ x;t) tan(x;t):
But
tan(x;t) = lim
x!0u
x=ux(x;t):
Likewise,
tan(x+ x;t) =ux(x+ x;t):
Hence, we get
Txutt(x;t) =ux(x+ x;t) ux(x;t):
Dividing by xand letting x!0 we obtain
Tutt(x;t) =uxx(x;t)
or
utt(x;t) =c2uxx(x;t) (13.2)
wherec2=T
:We callcthewave speed.
D'Alembert Solution of (13.2)
Letv=x+ctandw=x ct:Then by application of the chain rule we nd
ut=c(uv uw)
ux=uv+uw
utt=c2(uvv 2uvw+uww)
uxx=uvv+ 2uvw+uww:
Substituting into (13.2) we obtain
c2(uvv+ 2uvw+uww) =c2(uvv 2uvw+uww)
and this simpli es to
4c2uvw= 0 oruvw= 0:
13 HYPERBOLIC TYPE: THE WAVE EQUATION 107
It follows that
u(v;w) =f(v) +g(w)
wherefandgare arbitrary di erentiable functions. Now, writing uin terms
ofxandywe nd the general solution
u(x;y) =f(x+ct) +g(x ct):
D'Alembert's solution involves two arbitrary functions that are determined
(normally) by two initial conditions.
Example 13.1
Find the solution to the Cauchy problem
utt=c2uxx
u(x;0) =v(x)
ut(x;0) =w(x):
Solution.
We have
u(x;0) =f(x) +g(x) =v(x)
and
ut(x;0) =cf0(x) cg0(x) =w(x)
which implies that
f(x) g(x) =1
cW(x) =1
cZ
w(x)dx:
Therefore,
g(x) =1
2(v(x) 1
cW(x)):
Hence,
f(x) =1
2(v(x) +1
cW(x)):
Finally,
u(x;t) =1
2[v(x ct) +v(x+ct) +1
c(W(x+ct) W(x ct))]
=1
2[v(x ct) +v(x+ct) +1
cZx+ct
x ctw(s)ds]
108SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Practice Problems
Exercise 13.1
Show that if v(x;t) andw(x;t) satisfy equation (13.2) then v+ wis also
a solution to (13.2), where and are constants.
Exercise 13.2
Show that any linear time independent function u(x;t) =ax+bis a solution
to equation (13.2).
Exercise 13.3
Find a solution to (13.2) that satis es the homogeneous conditions u(x;0) =
u(0;t) =u(L;t) = 0:
Exercise 13.4
Solve the initial value problem
utt=9uxx
u(x;0) = cosx
ut(x;0) =0:
Exercise 13.5
Solve the initial value problem
utt=uxx
u(x;0) =1
1 +x2
ut(x;0) =0:
Exercise 13.6
Solve the initial value problem
utt=4uxx
u(x;0) =1
ut(x;0) = cos (2x):
13 HYPERBOLIC TYPE: THE WAVE EQUATION 109
Exercise 13.7
Solve the initial value problem
utt=25uxx
u(x;0) =v(x)
ut(x;0) =0
where
v(x) =1 ifx<0
0 ifx0:
Exercise 13.8
Solve the initial value problem
utt=c2uxx
u(x;0) =e x2
ut(x;0) = cos2x:
Exercise 13.9
Prove that the wave equation, utt=c2uxxsatis es the following properties,
which are known as invariance properties. If u(x;t) is a solution, then
(i) Any translate, u(x y;t) whereyis a xed constant, is also a solution.
(ii) Any derivative, say ux(x;t);is also a solution.
(iii) Any dilation, u(ax;at );is a solution, for any xed constant a.
Exercise 13.10
Findv(r) ifu(r;t) =v(r)
rcosntis a solution to the PDE
urr+2
rur=utt:
110SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Sample Exam Questions
Exercise 13.11
Find the solution of the wave equation on the real line ( 1< x < +1)
with the initial conditions
u(x;0) =ex; ut(x;0) = sinx:
Exercise 13.12
The total energy of the string (the sum of the kinetic and potential energies)
is de ned as
E(t) =1
2ZL
0(u2
t+c2u2
x)dx:
(a) Using the wave equation derive the equation of conservation of energy
dE(t)
dt=c2(ut(L;t)ux(L;t) ut(0;t)ux(0;t)):
(b) Assuming xed ends boundary conditions, that is the ends of the string
are xed so that u(0;t) =u(L;t) = 0;for allt>0;show that the energy is
constant.
(c) Assuming free ends boundary conditions for both x= 0 andx=L;that
is bothu(0;t) andu(L;t) vary with t;show that the energy is constant.
Exercise 13.13
For a wave equation with damping
utt c2uxx+dut= 0; d> 0;0<x<L
with the xed ends boundary conditions show that the total energy decreases.
Exercise 13.14
(a) Verify that for any twice di erentiable R(x) the function
u(x;t) =R(x ct)
is a solution of the wave equation utt=c2uxx:Such solutions are called
traveling waves.
(b) Show that the potential and kinetic energies (see Exercise 13.12) are
equal for the traveling wave solution in (a).
13 HYPERBOLIC TYPE: THE WAVE EQUATION 111
Exercise 13.15
Find the solution of the Cauchy wave equation
utt= 4uxx
u(x;0) =x2; ut(x;0) = sin 2x:
Simplify your answer as much as possible.
112SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
14 Parabolic Type: The Heat Equation in One-
Dimensional Space
In this section, We will look at a model for describing the distribution of
temperature in a solid material as a function of time and space.
Before we begin our discussion of the mathematics of the heat equation, we
must rst determine what is meant by the term heat ? Heat is type of energy
known as thermal energy . Heat travels in waves like other forms of energy,
and can change the matter it touches. It can heat it up and cause chemical
reactions like burning to occur.
Heat can be released through a chemical reaction (such as the nuclear re-
actions that make the Sun \burn") or can be trapped for a limited time by
insulators. It is often released along with other kinds of energy such as light
waves or sound waves. For example, a burning candle releases light and heat
waves. On the other hand, an explosion releases light, heat, and sound waves.
The most common units of heat are BTU (British Thermal Unit), Calorie
and Joule.
Consider now a rod made of homogeneous heat conducting material (i.e. it
is composed of the exact same material and no foreign bodies are in it) of
uniform density and constant cross section A;placed along the x axis from
x= 0 tox=Las shown in Figure 14.1.
Figure 14.1
Assume the heat
ows only in the x direction, with the lateral sides well
insulated, and the only way heat can enter or leave the rod is at either end.
Also we assume that the temperature of the rod is constant at any point of
the cross section. In other words, temperature will only vary in xand we
can hence consider the rod to be a one spatial dimensional rod. We will also
assume that heat energy in any piece of the rod is conserved.
Letu(x;t) be the temperature of the cross section at the point xand the
timet:Consider a portion Uof the rod from xtox+ xof length xas
14 PARABOLIC TYPE: THE HEAT EQUATION IN ONE-DIMENSIONAL SPACE 113
shown in Figure 14.2.
Figure 14.2
Consider the portion SofUof height s:From the theory of heat conduction,
the quantity of heat Qfromxtox+ sat timetis given by
Q=cu(x;t)V
where Vis the volume of Sandcis the speci c heat , that is, the amount
of heat energy that it takes to raise one unit of mass of the material by one
unit of temperature.
ButSis a cylinder of height sand area of base Aso that V=As:
Hence,
Q=cAu (x;t)s:
The quantity of heat in the portion Uis given by
Q(t) =Zx+x
xcAu (s;t)ds:
By di erentiating we take the partial of uto nd the change in heat with
respect to time.
dQ
dt=Zx+x
xcAut(s;t)ds:
Assuming that uis continuously di erentiable, we can apply the mean value
theorem for integrals and nd xx+ xsuch that
Zx+x
xut(s;t)ds= xut(;t):
Thus, the rate of change of heat in Uis given by
dQ
dt=cAxut(;t):
114SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
On the other hand, by Fourier (or Fick's) law of heat conduction, the rate
of heat
ow through any cross section is proportional to the area Aand
the negative gradient of the temperature normal to the cross section, and
heat
ows in the direction of decreasing temperature. Thus, the rate of heat
owing inUthrough the cross section at xis KAux(x;t) and the rate of
heat
owing out of Uthrough the cross section at x+xis KAux(x+x;t);
whereKis the thermal conductivity of the rod.
Now, the conservation of energy law states
rate of change of heat in U= rate of heat
owing in rate of heat
owing
out
or mathematically written as,
cAxut(;t) = KAux(x;t) +KAux(x+ x;t)
or
cAxut(;t) =KA[ux(x+ x;t) ux(x;t)]:
Dividing this last equation by cAxand letting x!0 we obtain
ut(x;t) =kuxx(x;t) (14.1)
wherek=K
cis called the di usivity constant.
Equation (14.1) is the one dimensional heat equation which is second order,
linear, homogeneous, and of parabolic type.
The non-homogeneous heat equation
ut=kuxx+f(x)
is known as the heat equation with an external heat source f(x):An ex-
ample of an exterenal heat source is the heat generated from a candle placed
under the bar.
The function
E(t) =ZL
0cu(x;t)dx
is called the total thermal energy at timetof the entire rod.
Example 14.1
The two ends of a uniform rod of length Lare insulated. There is a con-
stant source of thermal energy q06= 0 and the temperature is initially
14 PARABOLIC TYPE: THE HEAT EQUATION IN ONE-DIMENSIONAL SPACE 115
u(x;0) =f(x):
(a) Write the equation and the boundary conditions for this model.
(b) Calculate the total thermal energy of the entire rod.
Solution.
(a) The model is given by the PDE
cut(x;t) =Kuxx+q0
with boundary conditions
ux(0;t) =ux(L;t) = 0:
(b) First note that
d
dtZL
0cu(x;t)dx=ZL
0cut(x;t)dx=ZL
0Kuxxdx+ZL
0q0dx
=KuxjL
0+q0L=q0L
sinceux(0;t) =ux(L;t) = 0:Integrating in time from 0 to twe nd
E(t) =q0Lt+C:
ButC=E(0) =RL
0cu(x;0)dx=RL
0cf(x)dx:Hence, the total thermal
energy is given by
E(t) =ZL
0cf(x)dx+q0Lt
Initial Boundary Value Problems
In order to solve the heat equation we must give the problem some initial
conditions. If you recall from the theory of ODE, the number of conditions
required for solving initial value problems always matched the highest order
of the derivative in the equation.
In partial di erential equations the same idea holds except now we have to
pay attention to the variable we are di erentiating with respect to as well.
So, for the heat equation we have got a rst order time derivative and so we
will need one initial condition and a second order spatial derivative and so
we will need two boundary conditions.
116SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
For the initial condition, we de ne the temperature of every point along the
rod at time t= 0 by
u(x;0) =f(x)
wherefis a given (prescribed) function of x:This function is known as the
initial temperature distribution.
The boundary conditions will tell us something about what the temperature
is doing at the ends of the bar. The conditions are given by
u(0;t) =T0andu(L;t) =TL:
and they are called as the Dirichlet conditions. In this case, the general
form of the heat equation initial boundary value problem is to nd u(x;t)
satisfying
ut(x;t) =kuxx(x;t);0xL; t> 0
u(x;0) =f(x);0xL
u(0;t) =T0;u(L;t) =TL; t> 0:
In the case of insulated endpoints, i.e. there is no heat
ow out of them, we
use the boundary conditions
ux(0;t) =ux(L;t) = 0:
These conditions are examples of what is known as Neumann boundary
conditions. In this case, the general form of the heat equation initial bound-
ary value problem is to nd u(x;t) satisfying
ut(x;t) =kuxx(x;t);0xL; t> 0
u(x;0) =f(x);0xL
ux(0;t) =ux(L;t) = 0; t> 0:
14 PARABOLIC TYPE: THE HEAT EQUATION IN ONE-DIMENSIONAL SPACE 117
Practice Problems
Exercise 14.1
Show that if u(x;t) andv(x;t) satisfy equation (14.1) then u+ vis also a
solution to (14.1), where and are constants.
Exercise 14.2
Show that any linear time independent function u(x;t) =ax+bis a solution
to equation (14.1).
Exercise 14.3
Find a linear time independent solution uto (14.1) that satis es u(0;t) =T0
andu(L;T) =TL:
Exercise 14.4
Show that to solve (14.1) with the boundary conditions u(0;t) =T0and
u(L;t) =TLit suces to solve (14.1) with the homogeneous boundary
conditions u(0;t) =u(L;t) = 0:
Exercise 14.5
Find a solution to (14.1) that satis es the conditions u(x;0) =u(0;t) =
u(L;t) = 0:
Exercise 14.6
Let (I) denote equation (14.1) together with intial condition u(x;0) =f(x),
wherefis not the zero function, and the homogeneous boundary conditions
u(0;t) =u(L;t) = 0:Suppose a nontrivial solution to (I) can be written in
the formu(x;t) =X(x)T(t):Show thatXandTsatisfy the ODE
X00
kX= 0 andT0 T= 0
for some constant :
Exercise 14.7
Consider again the solution u(x;t) =X(x)T(t):Clearly,T(t) =T(0)et:
Suppose that >0:
(a) Show that X(x) =Aexp +Be xp ;where =
kandAandBare
arbitrary constants.
(b) Show that AandBsatisfy the two equations A+B= 0 andA(eLp
e Lp ) = 0:
(c) Show that A= 0 leads to a contradiction.
(d) Using (b) and (c) show that eLp =e Lp :Show that this equality leads
to a contradiction. We conclude that <0:
118SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Exercise 14.8
Consider the results of the previous exercise.
(a) Show that X(x) =c1cos x+c2sin xwhere =q
k:
(b) Show that =n= kn22
L2;wherenis an integer.
Exercise 14.9
Show thatu(x;t) =Pn
k=1uk(x;t);whereun(x;t) =cne kn22
L2tsin n
L
xsat-
is es (14.1) and the homogeneous boundary conditions.
Exercise 14.10
Suppose that a wire is stretched between 0 and a:Describe the boundary
conditions for the temperature u(x;t) when
(i) the left end is kept at 0 degrees and the right end is kept at 100 degrees;
and
(ii) when both ends are insulated.
Exercise 14.11
Letut=uxxfor 0< x < andt >0 with boundary conditions u(0;t) =
0 =u(;t) and initial condition u(x;0) =f(x):LetE(t) =R
0(u2
t+u2
x)dx:
Show thatE0(t)<0:
Exercise 14.12
Suppose
ut=uxx+ 4; ux(0;t) = 5; ux(L;t) = 6; u(x;0) =f(x):
Calculate the total thermal energy of the one-dimensional rod (as a function
of time).
14 PARABOLIC TYPE: THE HEAT EQUATION IN ONE-DIMENSIONAL SPACE 119
Sample Exam Questions
Exercise 14.13
Consider the heat equation
ut=kuxx
forx2(0;1) andt>0;with boundary conditions u(0;t) = 2 andu(1;t) = 3
fort >0 and initial conditions u(x;0) =xforx2(0;1):A function v(x)
that satis es the equation v00(x) = 0;with conditions v(0) = 2 and v(1) = 3
is called a steady-state solution. That is, the steady-state solutions of the
heat equation are those solutions that don't depend on time. Find v(x):
Exercise 14.14
Consider the equation for the one-dimensional rod of length Lwith given
heat energy source:
ut=uxx+q(x):
Assume that the initial temperature distribution is given by u(x;0) =f(x):
Find the equilibrium (steady state) temperature distribution in the following
cases.
(a)q(x) = 0;u(0) = 0;u(L) =T:
(b)q(x) = 0;ux(0) = 0;u(L) =T:
(c)q(x) = 0;u(0) =T;ux(L) = :
Exercise 14.15
Consider the equation for the one-dimensional rod of length Lwith insulated
ends:
cut=Kuxx; ux(0;t) =ux(L;t) = 0:
(a) Give the expression for the total thermal energy of the rod.
(b) Show using the equation and the boundary conditions that the total
thermal energy is constant.
Exercise 14.16
Suppose
ut=uxx+x; u (x;0) =f(x); ux(0;t) = ; ux(L;t) = 7:
(a) Calculate the total thermal energy of the one-dimensional rod (as a func-
tion of time).
(b) From part (a) nd the value of for which a steady-state solution exist.
(c) For the above value of nd the steady state solution.
120SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
15 An Introduction to Fourier Series
In this and the next section we will have a brief look to the subject of Fourier
series. The point here is to do just enough to allow us to do some basic so-
lutions to partial di erential equations later in the book.
Motivation: In Calculus we have seen that certain functions may be repre-
sented as power series by means of the Taylor expansions. These functions
must have in nitely many derivatives, and the series provide a good approx-
imation only in some (often small) vicinity of a reference point.
Fourier series constructed of trigonometric rather than power functions, and
can be used for functions not only not di erentiable, but even discontinuous
at some points. The main limitation of Fourier series is that the underlying
function should be periodic.
Recall from calculus that a function series is a series where the summands
are functions. Examples of function series include power series, Laurent se-
ries, Fourier series, etc.
Unlike series of numbers, there exist many types of convergence of series of
functions, namely, pointwise, uniform, etc. We say that a series of functionsP1
n=1fn(x)converges pointwise to a function fif and only if the sequence
of partial sums
Sn(x) =f1(x) +f2(x) ++fn(x)
converges pointwise to f:We write
1X
n=1fn(x) = lim
n!1Sn(x) =f(x):
Likewise, we say that a series of functionsP1
n=1fn(x)converges uniformly
to a function fif and only if the sequence of partial sums fSng1
n=1converges
uniformly to f:
In this section we introduce a type of series of functions known as Fourier
series . They are given by
f(x) =a0
2+1X
n=1h
ancosn
Lx
+bnsinn
Lxi
; LxL(15.1)
whereanandbnare called the Fourier coecients. The expression on the
right is called a trigonometric series. Note that we begin the series with
a0
2as opposed to simply a0to simplify the coecient formula for anthat we
15 AN INTRODUCTION TO FOURIER SERIES 121
will derive later in this section.
The main questions we want to consider next are the questions of determin-
ing which functions can be represented by Fourier series and if so how to
compute the coecients anandbn:
Before answering these questions, we look at some of the properties of Fourier
series.
Periodicity Property
Recall that a function fis said to be periodic with period T > 0 if
f(x+T) =f(x) for allx;x+Tin the domain of f:The smallest value
ofTfor whichfis periodic is called the fundamental period. A graph of
aT periodic function is shown in Figure 15.1.
Figure 15.1
For aT periodic function we have
f(x) =f(x+T) =f(x+ 2T) =:
Note that the de nite integral of a T periodic function is the same over any
interval of length T:By Exercise 15.1 below, if fandgare two periodic func-
tions with common period T;then the product fgand an arbitrary linear
combination c1f+c2gare also periodic with period T:It is an easy exercise
to show that the Fourier series (15.1) is periodic with fundamental period 2 L:
Orthogonality Property
Recall from Calculus that for each pair of vectors ~ uand~ vwe associate a
scalar quantity ~ u~ vcalled the dot product of~ uand~ v:We say that ~ uand~ v
areorthogonal if and only if ~ u~ v= 0:We want to de ne a similar concept
for functions.
Letfandgbe two functions with domain the closed interval [ a;b]:We de ne
122SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
a function that takes a pair of functions to a scalar. Symbolically, we write
<f;g> =Zb
af(x)g(x)dx:
We call< f;g > theinner product offandg:We say that fandg
areorthogonal if and only if < f;g > = 0:A set of functions is said to
bemutually orthogonal if each distinct pair of functions in the set is
orthogonal.
Example 15.1
Show that the set
1;cos n
Lx
;sin n
Lx
:n2N
is mutually orthogonal in
[ L;L]:
Solution.
We have ZL
L1cosn
Lx
dx=L
nh
sinn
LxiL
L= 0
andZL
L1sinn
Lx
dx= L
nh
cosn
LxiL
L= 0:
Now, forn6=mwe have
ZL
Lcosm
Lx
cosn
Lx
dx=1
2ZL
L
cos(m+n)
Lx
+ cos(m n)
Lx
dx
=1
2L
(m+n)sin(m+n)
Lx
+L
(m n)sin(m n)
LxL
L= 0
where we used the trigonometric identity
cosacosb=1
2[cos (a+b) + cos (a b)]:
In the exercises below, we show that
ZL
Lsinm
Lx
sinn
Lx
dx= 0
15 AN INTRODUCTION TO FOURIER SERIES 123
and ZL
Lcosm
Lx
sinn
Lx
dx= 0
The reason we care about these functions being orthogonal is because we will
exploit this fact to develop a formula for the coecients in our Fourier series.
Now, in order to answer the rst question mentioned earlier, that is, which
functions can be expressed as a Fourier series expansion, we need to intro-
duce some mathematical concepts.
A function f(x) is said to be piecewise continuous on [a;b] if it is contin-
uous in [a;b] execept possibly at nitely many points of discontinuity within
the interval [ a;b];and at each point of discontinuity, the right- and left-
handed limits of fexist. An example of a piecewise continuous function is
the function
f(x) =x 0x<1
x2 x1x2:
We will say that fispiecewise smooth in [a;b] if and only if f(x) as well
as its derivatives are piecewise continuous.
The following theorem, proven in more advanced books, ensures that a
Fourier decomposition can be found for any function which is piecewise
smooth.
Theorem 15.1
Letfbe a 2L-periodic function. If fis a piecewise smooth on [ L;L] then
for all points of discontinuity x2( L;L) we have
f(x ) +f(x+)
2=a0
2+1X
n=1h
ancosn
Lx
+bnsinn
Lxi
:
where as for points of continuity x2( L;L) we have
f(x) =a0
2+1X
n=1h
ancosn
Lx
+bnsinn
Lxi
:
Remark 15.1
(1) Almost all functions occurring in practice are piecewise smooth functions.
(2) Given a non-periodic function fon [ L;L]. The above theorem applies
to the periodic extension FoffwhereF(x+ 2nL) =f(x) (n2Z) and
F(x) =f(x) on [ L;L]:
124SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Convergence Results of Fourier Series
We list few of the results regarding the convergence of Fourier series:
(1) The type of convergence in the above theorem is pointwise convergence.
(2) The convergence is uniform for a continuous function fon [ L;L] such
thatf( L) =f(L):
(3) The convergence is uniform wheneverP1
n=1(janj2+jbnj2) is convergent.
(4) Iff(x) is periodic, continuous, and has a piecewise continuous derivative,
then the Fourier Series corresponding to fconverges uniformly to f(x) for
the entire real line.
(5) The convergence is uniform on any closed interval that does not contain
a point of discontinuity.
Euler-Fourier Formulas
Next, we will answer the second question mentioned earlier, that is, the ques-
tion of nding formulas for the coecients anandbn:These formulas for an
andbnare called Euler-Fourier formulas which we derive next. We will as-
sume that the RHS in (15.1) converges uniformly to f(x) on the interval
[ L;L]:Integrating both sides of (15.1) we obtain
ZL
Lf(x)dx=ZL
La0
2dx+ZL
L1X
n=1h
ancosn
Lx
+bnsinn
Lxi
dx:
Since the trigonometric series is assumed to be uniformly convergent, from
Section 2, we can interchange the order of integration and summation to
obtain
ZL
Lf(x)dx=ZL
La0
2dx+1X
n=1ZL
Lh
ancosn
Lx
+bnsinn
Lxi
dx:
But ZL
Lcosn
Lx
dx=L
nsinn
LxiL
L= 0
and likewise
ZL
Lsinn
Lx
dx= L
ncosn
LxiL
L= 0:
Thus,
a0=1
LZL
Lf(x)dx:
15 AN INTRODUCTION TO FOURIER SERIES 125
To nd the other Fourier coecients, we recall the results of Exercises 15.2
- 15.3 below.
ZL
Lcosn
Lx
cosm
Lx
dx=Lifm=n
0 ifm6=n
ZL
Lsinn
Lx
sinm
Lx
dx=Lifm=n
0 ifm6=n
ZL
Lsinn
Lx
cosm
Lx
dx= 0;8m;n:
Now, to nd the formula for the Fourier coecients amform> 0;we multiply
both sides of (15.1) by cos m
Lx
and integrate from LtoLto otbain
ZL
Lf(x) cosm
Lx
=ZL
La0
2cosm
Lx
dx+1X
n=1
anZL
Lcosn
Lx
cosm
Lx
dx
+bnZL
Lsinn
Lx
cosm
Lx
dx:
Hence,ZL
Lf(x) cosm
Lx
dx=amL
and therefore
am=1
LZL
Lf(x) cosm
Lx
dx:
Likewise, we can show that
bm=1
LZL
Lf(x) sinm
Lx
dx:
Example 15.2
Find the Fourier series expansion of
f(x) =0; x0
x; x> 0
on the interval [ ;]:
126SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Solution.
We have
a0=1
Z
f(x)dx=1
Z
0xdx=
2
an=1
Z
0xcosnxdx =1
xsinnx
n+cosnx
n2
0=( 1)n 1
n2
bn=1
Z
0xsinnxdx =1
xcosnx
n+sinnx
n2
0=( 1)n+1
n
Hence,
f(x) =
4+1X
n=1( 1)n 1
n2cos (nx) +( 1)n+1
nsin (nx)
Example 15.3
Apply Theorem 15.1 to the function in Example 15.2.
Solution.
LetFbe a periodic extension of fof period 2:Thus,f(x) =F(x) on the
interval [ ;]:Clearly,Fis a piecewise smooth function so that by the
previous thereom we can write
4+1X
n=1( 1)n 1
n2cos (nx) +( 1)n+1
nsin (nx)
=8
<
:
2; ifx=
f(x);if <x<
2; ifx=
Takingx=we have the identity
4+1X
n=1( 1)n 1
n2( 1)n=
2
which can be simpli ed to
1X
n=11
(2n 1)2=2
8:
This provides a method for computing an approximate value of
15 AN INTRODUCTION TO FOURIER SERIES 127
Remark 15.2
An example of a function that does not have a Fourier series representation
is the function f(x) =1
x2on [ L;L]:For example, the coecient a0for this
function does not exist. Thus, not every function can be written as a Fourier
series expansion.
The nal topic of discussion here is the topic of di erentiation and integration
of Fourier series. In particular we want to know if we can di erentiate a
Fourier series term by term and have the result be the Fourier series of the
derivative of the function. Likewise we want to know if we can integrate a
Fourier series term by term and arrive at the Fourier series of the integral of
the function. Answers to these questions are provided next.
Theorem 15.2
A Fourier series of a piecewise smooth functionfcan always be integrated
term by term and the result is a convergent in nite series that always con-
verges toRL
Lf(x)dxeven if the original series has jumps.
Theorem 15.3
A Fourier series of a continuous function f(x) can be di erentiated term by
term iff0(x) ispiecewise smooth. The result of the di erentiation is the
Fourier series of f0(x):
128SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Practice Problems
Exercise 15.1
Letfandgbe two functions with common domain Dand common period
T:Show that
(a)fgis periodic of period T:
(b)c1f+c2gis periodic of period T;wherec1andc2are real numbers.
Exercise 15.2
Show that for m6=nwe have
(a)RL
Lsin m
Lx
sin n
Lx
dx= 0 and
(b)RL
Lcos m
Lx
sin n
Lx
dx= 0:
Exercise 15.3
Compute the following integrals:
(a)RL
Lcos2 n
Lx
dx:
(b)RL
Lsin2 n
Lx
dx:
(c)RL
Lcos n
Lx
sin n
Lx
dx:
Exercise 15.4
Find the Fourier coecients of
f(x) =8
<
: ; x<0
; 0<x<
0; x = 0;
on the interval [ ;]:
Exercise 15.5
Find the Fourier series of f(x) =x2 1
2on the interval [ 1;1]:
Exercise 15.6
Find the Fourier series of the function
f(x) =8
<
: 1; 2<x<
0; <x<
1; <x< 2:
15 AN INTRODUCTION TO FOURIER SERIES 129
Exercise 15.7
Find the Fourier series of the function
f(x) =1 +x; 2x0
1 x;0<x2:
Exercise 15.8
Show thatf(x) =1
xis not piecewise continuous on [ 1;1]:
Exercise 15.9
Assume that f(x) is continuous and has period 2 L:Prove that
ZL
Lf(x)dx=ZL+a
L+af(x)dx
is independent of a2R:In particular, it does not matter over which interval
the Fourier coecients are computed as long as the interval length is 2 L:
[Remark: This result is also true for piecewise continuous functions].
Exercise 15.10
Consider the function f(x) de ned by
f(x) =1 0x<1
2 1x<3
and extended periodically with period 3 to Rso thatf(x+ 3) =f(x) for all
x:
(i) Find the Fourier series of f(x):
(ii) Discuss its limit: In particular, does the Fourier series converge pointwise
or uniformly to its limit, and what is this limit?
(iii) Plot the graph of f(x) and the limit of the Fourier series.
130SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Sample Exam Questions
Exercise 15.11
For the following functions f(x) on the interval L<x<L , determine the
coecients an; n= 0;1;2;andbn;n2Nof the Fourier series expansion.
(a)f(x) = 1:
(b)f(x) = 2 + sin x
L
:
(c)f(x) =1x0
0x>0:
(d)f(x) =x:
Exercise 15.12
Letf(t) be the function with period 2 de ned as
f(t) =2 if 0x
2
0 if
2<x2
f(t) has a Fourier series and that series is equal to
a0
2+1X
n=1(ancosnt+bnsinnt):
Finda3
b3.
Exercise 15.13
Letf(x) =x3on [ ;];extended periodically to all of R:Find the Fourier
coecients an; n= 1;2;3;:
Exercise 15.14
Letf(x) be the square wave function
f(x) = x<0
0x
extended periodically to all of R:To what value does the Fourier series of
f(x) converge when x= 0?
Exercise 15.15
(a) Find the Fourier series of
f(x) =1 x<0
2 0x
15 AN INTRODUCTION TO FOURIER SERIES 131
extended periodically to all of R:Simplify your coecients as much as pos-
sible.
(b) Use (a) to evaluate the seriesP1
n=1( 1)n+1
(2n 1):Hint: Evaluate the Fourier
series atx=
2:
132SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
16 Fourier Sines Series and Fourier Cosines Se-
ries
In this section we discuss some important properties of Fourier series when
the underlying function fis either even or odd.
A function fisodd if it satis es f( x) = f(x) for allxin the domain of
fwhereasfiseven if it satis es f( x) =f(x) for allxin the domain of f:
Now, we recall from Exercises (1.6)-(1.7) the following facts about even and
odd functions. If f(x) is even then
ZL
Lf(x)dx= 2ZL
0f(x)dx:
Iffis odd thenZL
Lf(x)dx= 0:
Using just these basic facts we can gure out some important properties of
the Fourier series we get for odd or even functions.
Example 16.1
Show the following
(a) Iffandgare either both even or both odd then fgis even.
(b) Iffis odd and gis even then fgis odd.
Solution.
(a) Suppose that both fandgare even. Then ( fg)( x) =f( x)g( x) =
f(x)g(x) = (fg)(x):That is,fgis even. Now, suppose that both fandg
are odd. Then ( fg)( x) =f( x)g( x) = [ f(x)][ g(x)] = (fg)(x):That
is,fgis even.
(b)fis odd and gis even. Then ( fg)( x) =f( x)g( x) = f(x)g(x) =
(fg)(x):That is,fgis odd
Example 16.2
(a) Find the value of the integralRL
Lf(x) sin n
Lx
dxwhenfis even.
(b) Find the value of the integralRL
Lf(x) cos n
Lx
dxwhenfis odd.
Solution.
(a) Since the function sin n
Lx
is odd andfis even, we have that f(x) sin n
Lx
16 FOURIER SINES SERIES AND FOURIER COSINES SERIES 133
is odd so that ZL
Lf(x) sinn
Lx
dx= 0
(b) Since the function cos n
Lx
is even and fis odd, we have that f(x) cos n
Lx
is odd so that ZL
Lf(x) cosn
Lx
dx= 0
Even and Odd Extensions
Letf: [0;L]!Rbe a piecewise smooth function. We de ne the odd
extension of this function on the interval LxLby
fodd(x) =8
<
:f(x) 0<xL
f( x) Lx<0
0x= 0:
This function will be odd on the interval [ L;L];and will be equal to f(x)
on the interval (0 ;L]:We can then further extend this function to the entire
real line by de ning it to be 2 Lperiodic. Let fodddenote this extension. We
note thatfoddis an odd function and piecewise smooth so that by Theorem
15.1 it possesses a Fourier series expansion, and from the fact that it is odd
all of thea0
ns are zero. Moreover, in the interval [0 ;L] we have
f(x) =1X
n=1bnsinn
Lx
: (16.1)
We call (16.1) the Fourier sine series off:
The coecients bnare given by the formula
bn=1
LZL
Lfoddsinn
Lx
dx=2
LZL
0foddsinn
Lx
dx
=2
LZL
0f(x) sinn
Lx
dx
sincefoddsin n
Lx
is an even function.
Likewise, we can de ne the even extension offon the interval LxL
by
feven(x) =f(x) 0xL
f( x) Lx<0:
134SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
We can then further extend this function to the entire real line by de ning
it to be 2Lperiodic. Let fevendenote this extension. Again, we note that
fevenis equal to the original function f(x) on the interval upon which f(x)
is de ned. Since fevenis piecewise smooth, by Theorem 15.1 it possesses a
Fourier series expansion, and from the fact that it is even all of the b0
ns are
zero. Moreover, in the interval [0 ;L] we have
f(x) =a0
2+1X
n=1ancosn
Lx
: (16.2)
We call (16.2) the Fourier cosine series off:The coecients anare given
by
an=2
LZL
0f(x) cosn
Lx
dx; n = 0;1;2;:
Example 16.3
Graph the odd and even extensions of the function f(x) =x;0x1:
Solution.
We havefodd(x) =xfor 1x1:The odd extension of fis shown in
Figure 16.1(a). Likewise,
feven(x) =x 0x1
x 1x<0:
The even extension is shown in Figure 16.1(b)
Figure 16.1
16 FOURIER SINES SERIES AND FOURIER COSINES SERIES 135
Example 16.4
Find the Fourier sine series of the function
f(x) =x; 0x
2
x;
2x:
Solution.
We have
bn=2
"Z
2
0xsinnxdx +Z
2( x) sinnxdx#
:
Using integration by parts we nd
Z
2
0xsinnxdx =h
x
ncosnxi
2
0+1
nZ
2
0cosnxdx
= cos (n=2)
2n+1
n2[sinnx]
2
0
= cos (n=2)
2n+sin (n=2)
n2
while
Z
2( x) sinnxdx =
( x)
ncosnx
2 1
nZ
2cosnxdx
=cos (n=2)
2n 1
n2[sinnx]
2
=cos (n=2)
2n+sin (n=2)
n2
Thus,
bn=4 sin (n=2)
n2;
and the Fourier sine series of f(x) is
f(x) =1X
n=14 sin (n=2)
n2sinnx=1X
n=14( 1)2n 1
(2n 1)2sin (2n 1)x
136SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Practice Problems
Exercise 16.1
Give an example of a function that is both even and odd.
Exercise 16.2
Graph the odd and even extensions of the function f(x) = 1;0x1:
Exercise 16.3
Graph the odd and even extensions of the function f(x) =L xfor 0x
L:
Exercise 16.4
Graph the odd and even extensions of the function f(x) = 1 +x2for 0
xL:
Exercise 16.5
Find the Fourier cosine series of the function
f(x) =x; 0x
2
x;
2x
Exercise 16.6
Find the Fourier cosine series of f(x) =xon the interval [0 ;]:
Exercise 16.7
Find the Fourier sine series of f(x) = 1 on the interval [0 ;]:
Exercise 16.8
Find the Fourier sine series of f(x) = cosxon the interval [0 ;]:
Exercise 16.9
Find the Fourier cosine series of f(x) =e2xon the interval [0 ;1]:
16 FOURIER SINES SERIES AND FOURIER COSINES SERIES 137
Sample Exam Questions
Exercise 16.10
For the following functions on the interval [0 ;L], nd the coecients bnof
the Fourier sine expansion.
(a)f(x) = sin 2
Lx
:
(b)f(x) = 1
(c)f(x) = cos
Lx
:
Exercise 16.11
For the following functions on the interval [0 ;L], nd the coecients anof
the Fourier cosine expansion.
(a)f(x) = 5 + cos
Lx
:
(b)f(x) =x
(c)
f(x) =1 0<xL
2
0L
2<xL
Exercise 16.12
Consider a function f(x);de ned on 0xL;which is even (symmetric)
aroundx=L
2:Show that the even coecients ( neven) of the Fourier sine
series are zero.
Exercise 16.13
Consider a function f(x);de ned on 0xL;which is odd around x=L
2:
Show that the even coecients ( neven) of the Fourier cosine series are zero.
Exercise 16.14
The Fourier sine series of f(x) = cos x
L
for 0xLis given by
cosx
L
=1X
n=1bnsinnx
L
; n2N
where
b1= 0; bn=2n
(n2 1)[1 + ( 1)n]:
Using term-by-term integration, nd the Fourier cosine series of sin nx
L
:
138SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Exercise 16.15
Consider the function
f(x) =1 0x<1
2 1x<2
(a) Sketch the even extension of f:
(b) Finda0in the Fourier series for the even extension of f:
(c) Findan(n= 1;2;) in the Fourier series for the even extension of f:
(d) Findbnin the Fourier series for the even extension of f:
(e) Write the Fourier series for the even extension of f:
17 SEPARATION OF VARIABLES FOR PDES 139
17 Separation of Variables for PDEs
Finding analytic solutions to PDEs is essentially impossible. Most of the
PDE techniques involve a mixture of analytic, qualitative and numeric ap-
proaches. Of course, there are some easy PDEs too. If you are lucky your
PDE has a solution with separable variables. In this section we discuss the
application of the method of separation of variables in the solution of PDEs.
In developing a solution to a partial di erential equation by separation of
variables, one assumes that it is possible to separate the contributions of
the independent variables into separate functions that each involve only one
independent variable. Thus, the method consists of the following steps
1. Factorize the (unknown) dependent variable of the PDE into a product of
functions, each of the factors being a function of one independent variable.
That is,
u(x;y) =X(x)Y(y):
2. Substitute into the PDE, and divide the resulting equation by X(x)Y(y):
3. Then the problem turns into a set of separated ODEs (one for X(x) and
one forY(y):)
4. The general solution of the ODEs is found, and boundary initial condi-
tions are imposed.
5.u(x;y) is formed by multiplying together X(x) andY(y):
We illustrate these steps in the next two examples.
Example 17.1
Find all the solutions of the form u(x;t) =X(x)T(t) of the equation
uxx ux=ut
Solution.
It is very easy to nd the derivatives of a separable function:
ux=X0(x)T(t);ut=X(x)T0(t) anduxx=X00(x)T(t)
this is basically a consequence of the fact that di erentiation with respect
toxseestas a constant, and vice versa. Now the equation uxx ux=ut
becomes
X00(x)T(t) X0(x)T(t) =X(x)T0(t):
140SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
We can separate variables further. Division by X(x)T(t) gives
X00(x) X0(x)
X(x)=T0(t)
T(t):
The expression on the LHS is a function of xwhereas the one on the RHS is
a function of tonly. They both have to be constant. That is,
X00(x) X0(x)
X(x)=T0(t)
T(t)=:
Thus, we have the following ODEs:
X00 X0 X= 0 andT0=T:
The second equation is easy to solve: T(t) =Cet:The rst equation is
solved via the characteristic equation !2 ! = 0;whose solutions are
!=1p
1 + 4
2:
If> 1
4then
X(x) =Ae1+p1+4
2x+Be1 p1+4
2x:
In this case,
u(x;t) =De1+p1+4
2xet+Ee1 p1+4
2xet:
If= 1
4then
X(x) =Aex
2+Bxex
2
and in this case
u(x;t) = (D+Ex)ex
2 t
4:
If< 1
4then
X(x) =Aex
2cos p
(1 + 4)
2x!
+Bex
2sin p
(1 + 4)
2x!
:
In this case,
u(x;t) =D0ex
2+tcos p
(1 + 4)
2x!
+B0ex
2+tsin p
(1 + 4)
2x!
17 SEPARATION OF VARIABLES FOR PDES 141
Example 17.2
Solve Laplace's equation using the separation of variables method
u=uxx+uyy= 0:
Solution.
We look for a solution of the form u(x;y) =X(x)Y(y):Substituting in the
Laplace's equation, we obtain
X00(x)Y(y) +X(x)Y00(y) = 0:
AssumingX(x)Y(y) is nonzero, dividing for X(x)Y(y) and subtractingY00(y)
Y(y)
from both sides, we nd:
X00(x)
X(x)= Y00(y)
Y(y):
The left hand side is a function of xwhile the right hand side is a function
ofy:This says that they must equal to a constant. That is,
X00(x)
X(x)= Y00(y)
Y(y)=
whereis a constant. This results in the following two ODEs
X00 X= 0 andY00+Y= 0:
The solutions of these equations depend on the sign of :
If>0 then the solutions are given
X(x) =Aep
x+Be p
x
Y(y) =Ccos (p
y) +Dsin (p
y)
whereA;B;C; andDare constants. In this case,
u(x;t) =k1ep
xcos (p
y) +k2ep
xsin (p
y)
+k3e p
xcos (p
y) +k4e p
xsin (p
y):
If= 0 then
X(x) =Ax+B
Y(y) =Cy+D
142SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
whereA;B; andCare arbitrary constants. In this case,
u(x;y) =k1xy+k2x+k3y+k4:
If<0 then
X(x) =Acos (p
x) +Bsin (p
x)
Y(y) =Cep
y+De p
y
whereA;B;C; andDare arbitrary constants. In this case,
u(x;y) =k1cos (p
x)ep
y+k2cos (p
x)e p
y
+k3sin (p
x)ep
y+k4sin (p
x)e p
y
Example 17.3
Solve using the separation of variables method.
yux xuy= 0:
Solution.
Substituteu(x;y) =X(x)Y(y) into the given equation we nd
yX0Y xXY0= 0:
This can be separated into
X0
xX=Y0
yY:
The left hand side is a function of xwhile the right hand side is a function
ofy:This says that they must equal to a constant. That is,
X0
xX=Y0
yY=
whereis a constant. This results in the following two ODEs
X0 xX = 0 andY0 yY = 0:
Solving these equations using the method of separation of variable for ODEs
we ndX(x) =Aex2
2andY(y) =Bey2
2:Thus, the general solution is given
by
u(x;y) =Ce(x2+y2)
2
17 SEPARATION OF VARIABLES FOR PDES 143
Practice Problems
Exercise 17.1
Solve using the separation of variables method
u+u= 0:
Exercise 17.2
Solve using the separation of variables method
ut=kuxx:
Exercise 17.3
Derive the system of ordinary di erential equations for R(r) and () that
is satis ed by solutions to
urr+1
rur+1
r2u= 0:
Exercise 17.4
Derive the system of ordinary di erential equations and boundary conditions
forX(x) andT(t) that is satis ed by solutions to
utt=uxx 2u;0<x< 1; t> 0
u(0;t) = 0 =u(1;t)t>0
of the form u(x;t) =X(x)T(t):(Note: you do not need to solve for Xand
T.)
Exercise 17.5
Derive the system of ordinary di erential equations and boundary conditions
forX(x) andT(t) that is satis ed by solutions to
ut=kuxx;0<x<L; t> 0
u(x;0) =f(x); ux(0;t) = 0 =ux(L;t)t>0
of the form u(x;t) =X(x)T(t):(Note: you do not need to solve for Xand
T.)
Exercise 17.6
Find all product solutions of the PDE ux+ut= 0:
144SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Exercise 17.7
Derive the system of ordinary di erential equations for X(x) andY(y) that
is satis ed by solutions to
3uyy 5uxxxy+ 7uxxy= 0:
of the form u(x;y) =X(x)Y(y):
Exercise 17.8
Find the general solution by the method of separation of variables.
uxy+u= 0:
Exercise 17.9
Find the general solution by the method of separation of variables.
ux yuy= 0:
17 SEPARATION OF VARIABLES FOR PDES 145
Sample Exam Questions
Exercise 17.10
Find the general solution by the method of separation of variables.
utt uxx= 0:
Exercise 17.11
For the following PDEs nd the ODEs implied by the method of separation
of variables.
(a)ut=kr(rur)r
(b)ut=kuxx u
(c)ut=kuxx aux
(d)uxx+uyy= 0
(e)ut=kuxxxx:
Exercise 17.12
Find all solutions to the following partial di erential equation that can be
obtained via the separation of variables.
ux uy= 0:
Exercise 17.13
Separate the PDE uxx uy+uyy=uinto two ODEs with a parameter. You
do not need to solve the ODEs.
146SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
18 Solutions of the Heat Equation by the Sep-
aration of Variables Method
In this section we apply the method of separation of variables in solving the
one spatial dimension of the heat equation.
The Heat Equation with Dirichlet Boundary Conditions
Consider the problem of nding all nontrivial solutions to the heat equation
ut=kuxxthat satis es the initial time condition u(x;0) =f(x) and the
Dirichlet boundary conditions u(0;t) =T0andu(L;t) =TL:
From Exercise 14.4, it suces to solve the problem with the Dirichlet bound-
ary conditions being replaced by the homogeneous boundary conditions u(0;t) =
u(L;t) = 0 (that is, the endpoints are assumed to be at zero temperature)
withunot the trivial solution. Let's assume that the solution can be writ-
ten in the form u(x;t) =X(x)T(t):Substituting into the heat equation we
obtain
X00
X=T0
kT:
Since the LHS only depends on xand the RHS only depends on t;there must
be a constant such that
X00
X=andT0
kT=:
This gives the two ordinary di erential equations
X00 X= 0 andT0 kT = 0:
As far as the boundary conditions, we have
u(0;t) = 0 =X(0)T(t) =)X(0) = 0
and
u(L;t) = 0 =X(L)T(t) =)X(L) = 0:
Note thatTis not the zero function for otherwise u0 and this contradicts
our assumption that uis the non-trivial solution.
Next, we consider the three cases of the sign of :
18 SOLUTIONS OF THE HEAT EQUATION BY THE SEPARATION OF VARIABLES METHOD 147
Case 1:= 0
In this case, X00= 0:Solving this equation we nd X(x) =ax+b:Since
X(0) = 0 we nd b= 0:SinceX(L) = 0 we nd a= 0:Hence,X0 and
u(x;t)0:That is,uis the trivial solution.
Case 2:>0
In this case, X(x) =Aep
x+Be p
x:Again, the conditions X(0) =X(L) =
0 implyA=B= 0 and hence the solution is the trivial solution.
Case 3:<0
In this case, X(x) =Acosp
x+Bsinp
x:The condition X(0) = 0
impliesA= 0:The condition X(L) = 0 implies Bsinp
L= 0:We must
haveB6= 0 otherwise X(x) = 0 and this leads to the trivial solution. Since
B6= 0, we obtain sinp
L= 0 orp
L=nwheren2N:Solving for
we nd= n22
L2:Thus, we obtain in nitely many solutions given by
Xn(x) =Ansinn
Lx; n2N:
Now, solving the equation
T0 kT = 0
by the method of separation of variables we obtain
Tn(t) =Bne n22
L2kt; n2N:
Hence, the functions
un(x;t) =Cnsinn
Lx
e n22
L2kt; n2N
satisfyut=kuxxand the boundary conditions u(0;t) =u(L;t) = 0:
Now, in order for these solutions to satisfy the initial value condition u(x;0) =
f(x);we invoke the superposition principle of linear PDE to write
u(x;t) =1X
n=1Cnsinn
Lx
e n22
L2kt: (18.1)
To determine the unknown constants Cnwe use the initial condition u(x;0) =
f(x) in (18.1) to obtain
f(x) =1X
n=1Cnsinn
Lx
:
148SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Since the right-hand side is the Fourier sine series of fon the interval [0 ;L];
the coecients Cnare given by
Cn=2
LZL
0f(x) sinn
Lx
dx: (18.2)
Thus, the solution to the heat equation is given by (18.1) with the C0
ns
calculated from (18.2).
Remark 18.1
According to Exercise 14.4, the solution to the heat equation with non-
homogeneous condition u(0;t) =T0andu(L;t) =TLis given by
u(x;t) =1X
n=1Cnsinn
Lx
e n22
L2kt+T0+TL T0
Lx:
The Heat Equation with Neumann Boundary Conditions
When both ends of the bar are insulated, that is, there is no heat
ow out
of them, we use the boundary conditions
ux(0;t) =ux(L;t) = 0:
In this case, the general form of the heat equation initial boundary value
problem is to nd u(x;t) satisfying
ut(x;t) =kuxx(x;t);0xL; t> 0
u(x;0) =f(x);0xL
ux(0;t) =ux(L;t) = 0; t> 0:
Since 0 =ux(0;t) =X0(0)T(t) we obtain X0(0) = 0:Likewise, 0 = ux(L;t) =
X0(L)T(t) impliesX0(L) = 0:Now, di erentiating X(x) =Acosp
x+
Bsinp
xwith respect to xwe nd
X0(x) = p
Asinp
x+p
Bcosp
x:
The conditions X0(0) =X0(L) = 0 implyp
B= 0 andp
Asinp
L=
0:Hence,B= 0 and= n22
L2and
Xn(x) =Ancosn
Lx
; n= 0;1;2;
18 SOLUTIONS OF THE HEAT EQUATION BY THE SEPARATION OF VARIABLES METHOD 149
and
un(x;t) =Cncosn
Lx
e n22
L2kt:
By the superposition principle, the required solution to the heat equation
with Neumann boundary conditions is given by
u(x;t) =C0
2+1X
n=1Cncosn
Lx
e n22
L2kt
where
Cn=2
LZL
0f(x) cosn
Lx
dx; n2N:
150SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Practice Problems
Exercise 18.1
Find the temperature in a bar of length 2 whose ends are kept at zero
and lateral surface insulated if the initial temperature is f(x) = sin
2x
+
3 sin 5
2x
:
Exercise 18.2
Find the temperature in a homogeneous bar of heat conducting material of
lengthLwith its end points kept at zero and initial temperature distribution
given byf(x) =dx
L2(L x);0xL:
Exercise 18.3
Find the temperature in a thin metal rod of length L;with both ends insu-
lated (so that there is no passage of heat through the ends) and with initial
temperature in the rod f(x) = sin
Lx
:
Exercise 18.4
Solve the following heat equation with Dirichlet boundary conditions
ut=kuxx
u(0;t) =u(L;t) = 0
u(x;0) =1 0x<L
2
2L
2xL:
Exercise 18.5
Solve
ut=kuxx
u(0;t) =u(L;t) = 0
u(x;0) = 6 sin9
Lx
:
Exercise 18.6
Solve
ut=kuxx
subject to
ux(0;t) =ux(L;t) = 0
u(x;0) =0 0x<L
2
1L
2xL:
18 SOLUTIONS OF THE HEAT EQUATION BY THE SEPARATION OF VARIABLES METHOD 151
Exercise 18.7
Solve
ut=kuxx
subject to
ux(0;t) =ux(L;t) = 0
u(x;0) = 6 + 4 cos3
Lx
:
Exercise 18.8
Solve
ut=kuxx
subject to
ux(0;t) =ux(L;t) = 0
u(x;0) = 3 cos8
Lx
:
Exercise 18.9
Find the general solution u(x;t) of
ut=uxx u;0<x<L; t> 0
ux(0;t) = 0 =ux(L;t); t> 0:
Brie
y describe its behavior as t!1:
152SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Sample Exam Questions
Exercise 18.10 (Energy method)
Letu1andu2be two solutions to the Robin boundary value problem
ut=uxx u;0<x< 1; t> 0
ux(0;t) =ux(1;t) = 0; t> 0
u(x;0) =g(x);0<x< 1
De new(x;t) =u1(x;t) u2(x;t):
(a) Show that wsatis es the initial value problem
wt=wxx w;0<x< 1; t> 0
w(x;0) = 0;0<x< 1
(b) De ne E(t) =R1
0w2(x;t)dx0 for allt0:Show that E0(t)0:
Hence, 0E(t)E(0) for allt>0:
(c) Show that E(t) = 0; w(x;t) = 0:Hence, conclude that u1=u2:
Exercise 18.11
Consider the heat induction in a bar where the left end temperature is main-
tained at 0, and the right end is perfectly insulated. We assume k= 1 and
L= 1:
(a) Derive the boundary conditions of the temperature at the endpoints.
(b) Following the separation of variables approach, derive the ODEs for X
andT:
(c) Consider the equation in X(x):What are the values of X(0) andX(1)?
Show that solutions of the form X(x) = sinp
xsatisfy the ODE and one
of the boundary conditions. Can you choose a value of so that the other
boundary condition is also satis ed?
Exercise 18.12
Using the method of separation of variables nd the solution of the heat
equation
ut=kuxx
satisfying the following boundary and initial conditions:
(a)u(0;t) =u(L;t) = 0; u(x;0) = 6 sin 9x
L
(b)u(0;t) =u(L;t) = 0; u(x;0) = 3 sin x
L
sin 3x
L
18 SOLUTIONS OF THE HEAT EQUATION BY THE SEPARATION OF VARIABLES METHOD 153
Exercise 18.13
Using the method of separation of variables nd the solution of the heat
equation
ut=kuxx
satisfying the following boundary and initial conditions:
(a)ux(0;t) =ux(L;t) = 0; u(x;0) = cos x
L
+ 4 cos 5x
L
:
(b)ux(0;t) =ux(L;t) = 0; u(x;0) = 5:
Exercise 18.14
Find the solution of the following heat conduction partial di erential equation
ut= 8uxx;0<x< 4; t> 0
u(0;t) =u(4;t) = 0; t> 0
u(x;0) = 6 sinx;0<x< 4:
154SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
19 Elliptic Type: Laplace's Equations in Rect-
angular Domains
Boundary value problems are of great importance in physical applications.
Mathematically, a boundary-value problem consists of nding a function
which satis es a given partial di erential equation and particular bound-
ary conditions. Physically speaking, the problem is independent of time,
involving only space coordinates.
Just as initial-value problems are associated with hyperbolic PDE, bound-
ary value problems are associated with PDE of elliptic type. In contrast to
initial-value problems, boundary-value problems are considerably more di-
cult to solve.
The main model example of an elliptic type PDE is the Laplace equation
u=uxx+uyy= 0 (19.1)
where the symbol is referred to as the Laplacian . Solutions of this equa-
tion are called harmonic functions.
Example 19.1
Show that, for all ( x;y)6= (0;0);u(x;y) = ln (x2+y2) is a harmonic function.
Solution.
We have
ux=2x
x2+y2
uxx=2y2 2x2
(x2+y2)2
uy=2y
x2+y2
uyy=2x2 2y2
(x2+y2)2
Plugging these expressions into the equation we nd uxx+uyy= 0:Hence,
u(x;y) is harmonic
The Laplace equation is arguably the most important di erential equation in
19 ELLIPTIC TYPE: LAPLACE'S EQUATIONS IN RECTANGULAR DOMAINS 155
all of applied mathematics. It arises in an astonishing variety of mathemati-
cal and physical systems, ranging through
uid mechanics, electromagnetism,
potential theory, solid mechanics, heat conduction, geometry, probability,
number theory, and on and on.
There are two main modi cations of the Laplace equation: the Poisson
equation (a non-homogeneous Laplace equation):
u=f(x;y)
and the eigenvalue problem (theHelmholtz equation ):
u=u; 2R:
Solving Laplace's Equation (19.1)
Note rst that both independent variables are spatial variables and each
variable occurs in a 2nd order derivative and so we will need two boundary
conditions for each variable a total of four boundary conditions.
Consider (19.1) in the rectangle
=f(x;y) : 0xa;0ybg
with the Dirichlet boundary conditions
u(0;y) =f1(y); u(a;y) =f2(y); u(x;0) =g1(x); u(x;b) =g2(x)
where 0xaand 0yb:
The separation of variables method is most successful when the boundary
conditions are homogeneous. Thus, solving the Laplace's equation in
re-
quires solving four initial boundary conditions problems, where in each prob-
lem three of the four conditions are homogeneous. The four problems to be
solved are
(I)8
<
:uxx+uyy= 0
u(0;y) =f1(y);
u(a;y) =u(x;0) =u(x;b) = 0(II)8
<
:uxx+uyy= 0
u(a;y) =f2(y);
u(0;y) =u(x;0) =u(x;b) = 0
(III)8
<
:uxx+uyy= 0
u(x;0) =g1(x);
u(0;y) =u(a;y) =u(x;b) = 0(IV)8
<
:uxx+uyy= 0
u(x;b) =g2(x);
u(0;y) =u(a;y) =u(x;0) = 0
156SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
If we letui(x;y),i= 1;2;3;4, denote the solution of each of the above
problems, then the solution to our original system will be
u(x;y) =u1(x;y) +u2(x;y) +u3(x;y) +u4(x;y):
In each of the above problems, we will apply separation of variables to (19.1)
and nd a product solution that will satisfy the di erential equation and the
three homogeneous boundary conditions. Using the Principle of Superposi-
tion we will nd a solution to the problem and then apply the nal boundary
condition to determine the value of the constant(s) that are left in the prob-
lem. The process is nearly identical in many ways to what we did when we
were solving the heat equation.
We will illustrate how to nd u(x;y) =u4(x;y):So let's assume that the so-
lution can be written in the form u(x;y) =X(x)Y(y):Substituting in (19.1),
we obtain
X00(x)Y(y) +X(x)Y00(y) = 0:
AssumingX(x)Y(y) is nonzero, dividing for X(x)Y(y) and subtractingY00(y)
Y(y)
from both sides, we nd:
X00(x)
X(x)= Y00(y)
Y(y):
The left hand side is a function of xwhile the right hand side is a function
ofy:This says that they must equal to a constant. That is,
X00(x)
X(x)= Y00(y)
Y(y)=
whereis a constant. This results in the following two ODEs
X00 X= 0 andY00+Y= 0:
As far as the boundary conditions, we have for all 0 xaand 0yb
u(0;y) = 0 =X(0)Y(y) =)X(0) = 0
u(a;y) = 0 =X(a)Y(y) =)X(a) = 0
u(x;0) = 0 =X(x)Y(0) =)Y(0) = 0
u(x;b) =g2(x) =X(x)Y(b):
19 ELLIPTIC TYPE: LAPLACE'S EQUATIONS IN RECTANGULAR DOMAINS 157
Note thatXandYare not the zero functions for otherwise u0 and this
contradicts our assumption that uis the non-trivial solution.
Consider the rst equation: since X00 X= 0 the solution depends on the
sign of:If= 0 thenX(x) =Ax+B:Now, the conditions X(0) =X(a) = 0
implyA=B= 0 and so u0:So assume that 6= 0:If > 0 then
X(x) =Aep
x+Be p
x:Now, the conditions X(0) =X(a) = 0; 6= 0
implyA=B= 0 and hence the solution is the trivial solution. Hence, in
order to have a nontrivial solution we must have <0:In this case,
X(x) =Acosp
x+Bsinp
x:
The condition X(0) = 0 implies A= 0:The condition X(a) = 0 implies
Bsinp
a= 0:We must have B6= 0 otherwise X(x) = 0 and this leads to
the trivial solution. Since B6= 0, we obtain sinp
a= 0 orp
a=n
wheren2N:Solving for we ndn= n22
a2:Thus, we obtain in nitely
many solutions given by
Xn(x) = sinn
ax; n2N:
Now, solving the equation
Y00+Y= 0
we obtain
Yn(y) =anep ny+bne p ny=Ancoshp
ny+Bnsinhp
ny; n2N:
Using the boundary condition Y(0) = 0 we obtain An= 0 for all n2N:
Hence, the functions
un(x;y) =Bnsinn
ax
sinhn
ay
; n2N
satisfy (19.1) and the boundary conditions u(0;y) =u(a;y) =u(x;0) = 0:
Now, in order for these solutions to satisfy the boundary value condition
u(x;b) =g2(x);we invoke the superposition principle of linear PDE to write
u(x;y) =1X
n=1Bnsinn
ax
sinhn
ay
: (19.2)
To determine the unknown constants Bnwe use the boundary condition
u(x;b) =g2(x) in (19.2) to obtain
g2(x) =1X
n=1
Bnsinhn
ab
sinn
ax
:
158SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Since the right-hand side is the Fourier sine series of g2on the interval [0 ;a];
the coecients Bnare given by
Bn=2
aZa
0g2(x) sinn
ax
dx
[sinhn
ab
] 1: (19.3)
Thus, the solution to the Laplace's equation is given by (19.1) with the B0
ns
calculated from (19.3).
Example 19.2
Solve 8
<
:uxx+uyy= 0
u(0;y) =f1(y);
u(a;y) =u(x;0) =u(x;b) = 0
Solution.
Assume that the solution can be written in the form u(x;y) =X(x)Y(y):
Substituting in (19.1), we obtain
X00(x)Y(y) +X(x)Y00(y) = 0:
AssumingX(x)Y(y) is nonzero, dividing for X(x)Y(y) and subtractingY00(y)
Y(y)
from both sides, we nd:
X00(x)
X(x)= Y00(y)
Y(y):
The left hand side is a function of xwhile the right hand side is a function
ofy:This says that they must equal to a constant. That is,
X00(x)
X(x)= Y00(y)
Y(y)=
whereis a constant. This results in the following two ODEs
X00 X= 0 andY00+Y= 0:
As far as the boundary conditions, we have for all 0 xaand 0yb
u(0;y) =f1(y) =X(0)Y(y)
u(a;y) = 0 =X(a)Y(y) =)X(a) = 0
19 ELLIPTIC TYPE: LAPLACE'S EQUATIONS IN RECTANGULAR DOMAINS 159
u(x;0) = 0 =X(x)Y(0) =)Y(0) = 0
u(x;b) = 0 =X(x)Y(b) =)Y(b) = 0
Note thatXandYare not the zero functions for otherwise u0 and this
contradicts our assumption that uis the non-trivial solution.
Consider the second equation: since Y00+Y= 0 the solution depends on the
sign of:If= 0 thenY(y) =Ay+B:Now, the conditions Y(0) =Y(b) = 0
implyA=B= 0 and so u0:So assume that 6= 0:If < 0 then
Y(y) =Aep
y+Be p
y:Now, the condition Y(0) =Y(b) = 0 imply
A=B= 0 and hence the solution is the trivial solution. Hence, in order to
have a nontrivial solution we must have >0:In this case,
Y(y) =Acosp
y+Bsinp
y:
The condition Y(0) = 0 implies A= 0:The condition Y(b) = 0 implies
Bsinp
b= 0:We must have B6= 0 otherwise Y(y) = 0 and this leads to
the trivial solution. Since B6= 0, we obtain sinp
b= 0 orp
b=nwhere
n2N:Solving for we ndn=n22
b2:Thus, we obtain in nitely many
solutions given by
Yn(y) = sinn
by
; n2N:
Now, solving the equation
X00 X= 0; > 0
we obtain
Xn(x) =anepnx+bne pnx=Ancoshn
bx
+Bnsinhn
bx
; n2N:
However, this is not really suited for dealing with the boundary condition
X(a) = 0:So, let's also notice that the following is also a solution.
Xn(x) =Ancoshn
b(x a)
+Bnsinhn
b(x a)
; n2N:
Now, using the boundary condition X(a) = 0 we obtain An= 0 for alln2N:
Hence, the functions
un(x;y) =Bnsinn
by
sinhn
b(x a)
; n2N
160SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
satisfy (19.1) and the boundary conditions u(a;y) =u(x;0) =u(x;b) = 0:
Now, in order for these solutions to satisfy the boundary value condition
u(0;y) =f1(y);we invoke the superposition principle of linear PDE to write
u(x;y) =1X
n=1Bnsinn
by
sinhn
b(x a)
: (19.4)
To determine the unknown constants Bnwe use the boundary condition
u(0;y) =f1(y) in (19.4) to obtain
f1(y) =1X
n=1
Bnsinh
n
ba
sinn
by
:
Since the right-hand side is the Fourier sine series of f1on the interval [0 ;b];
the coecients Bnare given by
Bn=2
bZb
0f1(y) sinn
by
dyh
sinh
n
bai 1
: (19.5)
Thus, the solution to the Laplace's equation is given by (19.4) with the B0
ns
calculated from (19.5)
Example 19.3
Solve
uxx+uyy= 0;0<x<L; 0<y<H
u(0;y) =u(L;y) = 0;0<y<H
u(x;0) =uy(x;0); u(x;H) =f(x);0<x<L:
Solution.
Using separation of variables we nd
X00
X= Y00
Y=:
We rst solve X00 X= 0 0<x<L
X(0) =X(L) = 0
We ndn= n22
L2and
Xn(x) = sinn
Lx; n2N:
19 ELLIPTIC TYPE: LAPLACE'S EQUATIONS IN RECTANGULAR DOMAINS 161
Next we need to solve
Y00+Y= 0 0<y<H
Y(0) Y0(0) = 0
The solution of the ODE is
Yn(y) =Ancoshn
Ly
+Bnsinhn
Ly
; n2N:
The boundary condition Y(0) Y0(0) = 0 implies
An Bnn
L= 0:
Hence,
Yn=Bnn
Lcoshn
Ly
+Bnsinhn
Ly
; n2N:
Using the superposition principle and the results above we have
u(x;y) =1X
n=1Bnsinn
Lxhn
Lcoshn
Ly
+ sinhn
Lyi
:
Substituting in the condition u(x;H) =f(x) we nd
f(x) =1X
n=1Bnsinn
Lxhn
Lcoshn
LH
+ sinhn
LHi
:
Recall the Fourier sine series of fon [0;L] given by
f(x) =1X
n=1Ansinn
Lx
where
An=2
LZL
0f(x) sinn
Lx
dx:
Thus, the general solution is given by
u(x;y) =1X
n=1Bnsinn
Lxhn
Lcoshn
Ly
+ sinhn
Lyi
:
with theBnsatisfying
Bnhn
Lcoshn
LH
+ sinhn
LHi
=2
LZL
0f(x) sinn
Lx
dx
162SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Practice Problems
Exercise 19.1
Solve 8
<
:uxx+uyy= 0
u(a;y) =f2(y);
u(0;y) =u(x;0) =u(x;b) = 0:
Exercise 19.2
Solve 8
<
:uxx+uyy= 0
u(x;0) =g1(x);
u(0;y) =u(a;y) =u(x;b) = 0:
Exercise 19.3
Solve 8
<
:uxx+uyy= 0
u(x;0) =u(0;y) = 0;
u(1;y) = 2y;u(x;1) = 3 sinx+ 2x
where 0x1 and 0y1:Hint: De ne U(x;y) =u(x;y) 2xy:
Exercise 19.4
Show thatu(x;y) =x2 y2andu(x;y) = 2xyare harmonic functions.
Exercise 19.5
Solve
uxx+uyy= 0;0xL; H
2yH
2
subject to
u(0;y) =u(L;y) = 0; H
2<y<H
2
u(x; H
2) =f1(x); u(x;H
2) =f2(x);0xL:
Exercise 19.6
Consider a complex valued function f(z) =u(x;y)+iv(x;y) wherei=p 1:
We say that fisholomorphic oranalytic if and only if fcan be expressed
as a power series in z;i.e.
u(x;y) +iv(x;y) =1X
n=0anzn:
(a) By di erentiating with respect to xandyshow that
19 ELLIPTIC TYPE: LAPLACE'S EQUATIONS IN RECTANGULAR DOMAINS 163
ux=vyanduy= vx
These are known as the Cauchy-Riemann equations.
(b) Show that u= 0 and v= 0:
Exercise 19.7
Show that Laplace's equation in polar coordinates is given by
urr+1
rur+1
r2u= 0:
Exercise 19.8
Solve
uxx+uyy= 0;0x2;0y3
subject to
u(x;0) = 0; u(x;3) =x
2
u(0;y) = sin4
3y
; u(2;y) = 7:
Exercise 19.9
Solve
uxx+uyy= 0;0xL;0yH
subject to
uy(x;0) = 0; u(x;H) = 0
u(0;y) =u(L;y) = 4 cosy
2H
:
Exercise 19.10
Solve
uxx+uyy= 0; x> 0;0yH
subject to
u(0;y) =f(y);ju(x;0)j<1
uy(x;0) =uy(x;H) = 0:
164SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Sample Exam Questions
Exercise 19.11
Consider Laplace's equation inside a rectangle
uxx+uyy= 0;0xL;0yH
subject to the boundary conditions
u(0;y) = 0; u(L;y) = 0; u(x;0) uy(x;0) = 0; u(x;H) = 20 sinx
L
5 sin3x
L
:
Find the solution u(x;y):
Exercise 19.12
Solve Laplace'e equation uxx+uyy= 0 in the rectangle 0 <x;y< 1 subject
to the conditions
u(0;y) =u(1;y) = 0;0<y< 1
u(x;0) = sin (2x); ux(x;0) = 2sin (2x);0<x< 1:
Exercise 19.13
Find the solution to Laplace's equation on the rectangle 0 <x< 1;0<y< 1
with boundary conditions
u(x;0) = 0; u(x;1) = 1
ux(0;y) =ux(1;y) = 0:
Exercise 19.14
Solve Laplace's equation on the rectangle 0 < x < a; 0< y < b with the
boundary conditions
ux(0;y) = a; ux(a;y) = 0
uy(x;0) =b; uy(x;b) = 0:
Exercise 19.15
Solve Laplace's equation on the rectangle 0 < x < ; 0< y < 2 with the
boundary conditions
u(0;y) =u(;y) = 0
uy(x;0) = 0; uy(x;2) = 2 sin 3x 5 sin 10x:
20 LAPLACE'S EQUATIONS IN CIRCULAR REGIONS 165
20 Laplace's Equations in Circular Regions
In the previous section we solved the Dirichlet problem for Laplace's equation
on a rectangular region. However, if the domain of the solution is a disc,
an annulus, or a circular wedge, it is useful to study the two-dimensional
Laplace's equation in polar coordinates.
It is well known in calculus that the cartesian coordinates ( x;y) and the polar
coordinates ( r;) of a point are related by the formulas
x=rcosandy=rsin
wherer= (x2+y2)1
2and tan=y
x:Using the chain rule we obtain
ux=urrx+ux= cosur sin
ru
uxx=uxrrx+uxx
=
cosurr+sin
r2u sin
rur
cos
+
sinur+ cosur cos
ru sin
ru
sin
r
uy=urry+uy= sinur+cos
ru
uyy=uyrry+uyy
=
sinurr cos
r2u+cos
rur
sin
+
cosur+ sinur sin
ru+cos
rucos
r
Substituting these equations into u= 0 we obtain
urr+1
rur+1
r2u= 0: (20.1)
Example 20.1
Find the solution to
u= 0; x2+y2<a2
subject to
(i) Boundary condition: u(a;) =f(); :
(ii) Boundedness at the origin: ju(0;)j<1:
(iii) Periodicity: u(r;+ 2) =u(r;); :
166SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Solution.
First, note that (iii) implies that u(r; ) =u(r;) andu(r; ) =u(r;):
Next, we will apply the method of separation of variables to (21.1). Suppose
that a solution u(r;) of (21.1) can be written in the form u(r;) =R(r)():
Substituting in (21.1) we obtain
R00(r)() +1
rR0(r)() +1
r2R(r)00() = 0:
Dividing by R (under the assumption that R6= 0) we obtain
00()
()= r2R00(r)
R(r) rR0(r)
R(r):
The left-hand side is independent of rwhereas the right-hand side is inde-
pendent of so that there is a constant such that
00()
()=r2R00(r)
R(r)+rR0(r)
R(r)=:
This results in the following ODEs
00() +() = 0 (20.2)
and
r2R00(r) +rR0(r) R(r) = 0: (20.3)
The second equation is known as Euler's equation. Both of these equations
are easily solvable. To solve (20.2), we only have to add the appropriate
boundary conditions. From (iii), we have ( ) = () and 0( ) =
0():If>0 then () =Acos (p
) +Bcos (p
):Using the condition
( ) = () we obtain 2 Bsin (p
) = 0:Using the condition 0( ) =
0() we obtain 2p
Asin (p
) = 0:If sin (p
)6= 0 thenA=B= 0
and we get the trivial solution. Therefore, we require sin (p
) = 0 and this
leads ton=n2forn= 1;2;:Note that we start with n= 1 since>0:
Hence,
n() =Ancosn+Bnsinn:
If= 0 then ( ) =A+Band the conditions ( ) = () and 0( ) =
0() leads to ( ) =A:If<0 then
() =Acosh (p
) +Bsinh (p
)
20 LAPLACE'S EQUATIONS IN CIRCULAR REGIONS 167
and applying the conditions ( ) = () and 0( ) = 0() we nd
A=B= 0:In summary, we have
n() =A0
ncosn+B0
nsinn;n = 0;1;2
The equation in Ris of Euler type and its solution must be of the form
R(r) =r :Substituting into (20.3) and using =n2;we nd
( 1)r + r n2r = 0:
Solving this equation we nd =n:Hence, we let
Rn(r) =Cnrn+Dnr n;n2N:
Forn= 0,R= 1 is a solution. To nd a second solution, we solve the
equation
r2R00+rR0= 0:
This can be done by dividing through by rand using the substitution S=R0
to obtainrS0+S= 0:Solving this by noting that the left-hand side is just
(rS)0we ndS=c
r:Hence,R0=c
rand this implies R(r) =Clnr:Thus,
R= 1 andR= lnrform a couple of linearly independent solutions of (20.3)
and so a general solution is given by
R0(r) =C0+D0lnr:
By assumption (ii), u(r;) must be bounded near r= 0;and so does Rn:
Sincer nand lnrare unbounded near r= 0, we must set D0=Dn= 0:In
this case, the solutions to Euler's equation are given by
Rn(r) =Cnrn; n= 0;1;2;:
Using the superposition principle, and combining the results obtained above,
we nd
u(r;) =C0+1X
n=1rn(Ancosn+Bnsinn):
Now, using the boundary condition u(a;) =f() we can write
f() =C0+1X
n=1(anAncosn+anBnsinn)
168SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
which is usually written in a more convenient equivalent form by
f() =a0
2+1X
n=1(ancosn+bnsinn):
It is obvious that anandbnare the Fourier coecients, and therefore can be
determined by the formulas
an=1
Z
f() cosnd; n = 0;1;
and
bn=1
Z
f() sinnd; n = 1;2;:
Finally, the general solution to our problem is given by
u(r;) =C0+1X
n=1rn(Ancosn+Bnsinn)
where
C0=a0
2=1
2Z
f()d
An=an
an=1
anZ
f() cosnd; n = 1;2;
Bn=bn
an=1
anZ
f() sinnd; n = 1;2;
Example 20.2
Find a 2 periodic function solution
u= 0; <; 1r2
subject to
u(1;) =u(2;) = sin; <:
Solution.
Use separation of variables. First, solving for ( ));we see that in order
to ensure that the solution is 2 periodic in ;the eigenvalues are =n2:
When solving the equation for R(r);we do NOT need to throw out solutions
20 LAPLACE'S EQUATIONS IN CIRCULAR REGIONS 169
which are not bounded as r!0:This is because we are working in the
annulus where ris bounded away from 0 and 1:Therefore, we obtain the
general solution
u(r;) = (C0+C1lnr)+1X
n=1[(Cnrn+Dnr n) cosn+(Anrn+Bnr n) sinn]:
But
C0+1X
n=1[(Cn+Dn) cosn+ (An+Bn) sinn] = sin
and
C0+C1ln 2 +1X
n=1[(Cn2n+Dn2 n) cosn+ (An2n+Bn2 n) sinn] = sin
Hence, comparing coecients we must have
C0=0
Cn+Dn=0
An+Bn=0n6= 1
A1+B1=1
Cn2n+Dn2 n=0
An2n+Bn2 n=0n6= 1
2A1+ 2 1B1=1
Solving these equations we nd C0=Cn=Dn= 0;A1=1
3;B1=2
3;and
An=Bn= 0 forn6= 1:Hence, the solution to the problem is
u(r;) =1
3
r+2
r
sin
Example 20.3
Solve Laplace's equation inside a 60wedge of radius asubject to the bound-
ary conditions:
(1)u(a;) =1
3cos 9 1
9cos 3:
(2)u(r;0) = 0; u(r;
3) = 0:
(3)ju(0;)j<1:
170SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Solution.
Lettingu(r;) =R(r)() and separating the variables we obtain the eigen-
value problem
00() +() = 0:
As above, one can easily see that the solution is of the form
() =Acosp
+Bsinp
:
The condition 0(0) = 0 implies B= 0:The condition 0
3
= 0 implies
n= (3n)2; n= 0;1;2;:Thus, the angular solution is
n() =A0
ncos 3n; n = 0;1;2;
The corresponding solutions of the radial problem are
Rn(r) =Anr3n+Bnr 3n; n= 0;1;:
To obtain a solution that remains bounded as r!0 we takeBn= 0:Hence,
un(r;) =1X
n=0Cnr3ncos 3n; n = 0;1;2;
Using the boundary condition
u(a;) =1
3cos 9 1
9cos 3
we obtainC1a3= 1
9andC3a9=1
3and 0 otherwise. Thus,
u(a;) =1
3r
a9
cos 9 1
9r
a3
cos 3
20 LAPLACE'S EQUATIONS IN CIRCULAR REGIONS 171
Practice Problems
Exercise 20.1
Solve the Laplace's equation in the unit disk with u(1;) = 3 sin 5:
Exercise 20.2
Solve the Laplace's equation in the upper half of the unit disk with u(1;) =
:
Exercise 20.3
Solve the Laplace's equation in the unit disk with ur(1;) = 2 cos 2:
Exercise 20.4
Consider
u(r;) =C0+1X
n=1rn(Ancosn+Bnsinn)
with
C0=a0
2=1
2Z
f( )d
An=an
an=1
anZ
f( ) cosn d ; n = 1;2;
Bn=bn
an=1
anZ
f( ) sinn d ; n = 1;2;
Using the trigonometric identity
cosacosb+ sinasinb= cos (a b)
show that
u(r;) =1
2Z
f( )"
1 + 21X
n=1r
an
cosn( )#
d :
Exercise 20.5
(a) Using Euler's formula from complex analysis eit= cost+isintshow that
cost=1
2(eit+e it);
172SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
wherei=p 1:
(b) Show that
1 + 21X
n=1r
an
cosn( ) = 1 +1X
n=1r
an
ein( )+1X
n=1r
an
e in( ):
(c) Letq1=r
aei( )andq2=r
ae i( ):It is de ned in complex analysis that
the absolute value of a complex number z=x+iyis given byjzj= (x2+y2)1
2:
Using these concepts, show that jq1j<1 andjq2j<1:
Exercise 20.6
(a)Show that
1X
n=1r
an
ein( )=rei( )
a rei( )
and1X
n=1r
an
e in( )=re i( )
a re i( )
Hint: Each sum is a geoemtric series with a ratio less than 1 in absolute
value so that these series converges.
(b) Show that
1 + 21X
n=1r
an
cosn( ) =a2 r2
a2 2arcos ( ) +r2:
Exercise 20.7
Show that
u(r;) =a2 r2
2Z
f( )
a2 2arcos ( ) +r2d :
This is known as the Poisson formula in polar coordinates.
Exercise 20.8
Solve
uxx+uyy= 0; x2+y2<1
subject to
u(1;) =; :
20 LAPLACE'S EQUATIONS IN CIRCULAR REGIONS 173
Exercise 20.9
The vibrations of a symmetric circular membrane where the displacement
u(r;t) depends on randtonly can be describe by the one-dimensional wave
equation in polar coordinates
utt=c2(urr+1
rur);0<r<a; t> 0
with initial condition
u(a;t) = 0; t> 0
and boundary conditions
u(r;0) =f(r); ut(r;0) =g(r);0<r<a:
(a) Show that the assumption u(r;t) =R(r)T(t) leads to the equation
1
c2T00
T=1
RR00+1
rR0
R=:
(b) Show that <0:
Exercise 20.10
Cartesian coordinates and cylindrical coordinates are shown in Figure 20.1
below.
Figure 20.1
174SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
(a) Show that x=rcos; y=rsin; z =z:
(b) Show that
uxx+uyy+uzz=urr+1
rur+1
r2u+uzz:
20 LAPLACE'S EQUATIONS IN CIRCULAR REGIONS 175
Sample Exam Questions
Exercise 20.11
An important result about harmonic functions is the so-called the maximum
principle which states: Any harmonic function u(x;y) de ned in a domain
satis es the inequality
min
(x;y)2@
u(x;y)u(x;y)max
(x;y)2@
u(x;y);8(x;y)2
[@
where@
denotes the boundary of
:
Letube harmonic in
= f(x;y) :x2+y2<1gand satis es u(x;y) = 2 x
for all (x;y)2@
:Show thatu(x;y)>0 for all (x;y)2
:
Exercise 20.12
Letube harmonic in
= f(x;y) :x2+y2<1gand satis es u(x;y) = 1 + 3x
for all (x;y)2@
:Determine
(i) max (x;y)2
u(x;y)
(ii) min (x;y)2
u(x;y)
without solving u= 0:
Exercise 20.13
Letu1(x;y) andu2(x;y) be harmonic functions on a smooth domain
such
that
u1j@
=g1(x;y) andu2j@
=g3(x;y)
whereg1andg2are continuous functions satisfying
max
(x;y)2@
g1(x;y)<min
(x;y)2@
g1(x;y):
Prove that u1(x;y)<u 2(x;y) for all (x;y)2
[@
:
Exercise 20.14
Show thatrncos (n) andrnsin (n) satisfy Laplace's equation in polar co-
ordinates.
Exercise 20.15
Solve the Dirichlet problem
u= 0;0r<a;
u(a;) = sin2:
176SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Exercise 20.16
Solve Laplace's equation
uxx+uyy= 0
outside a circular disk ( ra) subject to the boundary condition
u(a;) = ln 2 + 4 cos 3 :
You may assume that the solution remains bounded as r!1:
The Laplace Transform
Solutions for PDEs
If in a partial di erential equation the time tis one of the independent vari-
ables of the searched-for function, we say that the PDE is an evolution
equation. Examples of evolutions equations are the heat equation and the
wave equation. In contrast, when the equation involves only spatial indepen-
dent variables then the equation is called a stationary equation. Examples
of stationary equations are the Laplace's equations and Poisson equations.
There are classes of methods that can be used for solving the initial value or
initial boundary problems for evolution equations. We refer to these meth-
ods as the methods of integral transforms. The fundamental ones are the
Laplace and the Fourier transforms. In this chapter we will just consider the
Laplace transform.
21 Essentials of the Laplace Transform
Laplace transform has been introduced in an ODE course, and is used espe-
cially to solve linear ODEs with constant coecients, where the equations
are transformed to algebraic equations. This idea can be easily extended
to PDEs, where the transformation leads to the decrease of the number of
independent variables. PDEs in two variables are thus reduced to ODEs. In
this section we review the Laplace transform and its properties.
Laplace transform is yet another operational tool for solving constant coe-
cients linear di erential equations. The process of solution consists of three
main steps:
The given \hard" problem is transformed into a \simple" equation.
This simple equation is solved by purely algebraic manipulations.
The solution of the simple equation is transformed back to obtain the so-
177
178 THE LAPLACE TRANSFORM SOLUTIONS FOR PDES
lution of the given problem.
In this way the Laplace transformation reduces the problem of solving a dif-
ferential equation to an algebraic problem. The third step is made easier by
tables, whose role is similar to that of integral tables in integration.
The above procedure can be summarized by Figure 21.1
Figure 21.1
The Laplace transform is de ned in the following way. Let f(t) be de ned
fort0:Then the Laplace transform off;which is denoted by L[f(t)]
or byF(s), is de ned by the following equation
L[f(t)] =F(s) = lim
T!1ZT
0f(t)e stdt=Z1
0f(t)e stdt
The integral which de nes a Laplace transform is an improper integral. An
improper integral may converge ordiverge , depending on the integrand.
When the improper integral is convergent then we say that the function f(t)
possesses a Laplace transform. So what types of functions possess Laplace
transforms, that is, what type of functions guarantees a convergent improper
integral.
Example 21.1
Find the Laplace transform, if it exists, of each of the following functions
(a)f(t) =eat(b)f(t) = 1 (c)f(t) =t(d)f(t) =et2
Solution.
(a) Using the de nition of Laplace transform we see that
L[eat] =Z1
0e (s a)tdt= lim
T!1ZT
0e (s a)tdt:
But ZT
0e (s a)tdt=T ifs=a
1 e (s a)T
s aifs6=a:
21 ESSENTIALS OF THE LAPLACE TRANSFORM 179
For the improper integral to converge we need s>a: In this case,
L[eat] =F(s) =1
s a; s>a:
(b) In a similar way to what was done in part (a), we nd
L[1] =Z1
0e stdt= lim
T!1ZT
0e stdt=1
s; s> 0:
(c) We have
L[t] =Z1
0te stdt=
te st
s e st
s21
0=1
s2; s> 0:
(d) Again using the de nition of Laplace transform we nd
L[et2] =Z1
0et2 stdt:
Ifs0 thent2 st0 so thatet2 st1 and this implies thatR1
0et2 stdtR1
0dt:Since the integral on the right is divergent, by the comparison theorem
of improper integrals (see Theorem 23.1 below) the integral on the left is also
divergent. Now, if s >0 thenR1
0et(t s)dtR1
sdt:By the same reasoning
the integral on the left is divergent. This shows that the function f(t) =et2
does not possess a Laplace transform
The above example raises the question of what class or classes of functions
possess a Laplace transform. To answer this question we introduce few math-
ematical concepts.
A function fthat satis es
jf(t)jMeat; tC (21.1)
is said to be a function with an exponential order a. IfC= 0 in
(21.1) then the function is said to be exponentially bounded. Clearly,
if limt!1e atf(t) = 0 for some a>0 thenfis of exponential order a:
A function fis called piecewise continuous on an interval if the interval
can be broken into a nite number of subintervals on which the function is
continuous on each open subinterval (i.e. the subinterval without its end-
points) and has a nite limit at the endpoints ( jump discontinuities and
180 THE LAPLACE TRANSFORM SOLUTIONS FOR PDES
no vertical asymptotes) of each subinterval. Below is a sketch of a piecewise
continuous function.
Note that a piecewise continuous function is a function that has a nite
number of breaks in it and doesn't blow up to in nity anywhere. A function
de ned for t0 is said to be piecewise continuous on the in nite in-
terval if it is piecewise continuous on 0 tTfor allT > 0:Also, note
that a bounded continuous function is piecewise continuous.
Example 21.2
Show that the following functions are piecewise continuous and exponentially
bounded for t0:
(a)f(t) =tn(b)f(t) =tnsinat
Solution.
(a) Sinceet=P1
n=0tn
n!tn
n!; tnn!etfor allt0:Hence,tnis continuous
(and hence piecewise consitnuous) and exponentially bounded.
(b) Sincejtnsinatjn!et;tnsinatis piecewise continuous and exponentially
bounded
The following is an existence result of Laplace transform.
Theorem 21.1
Suppose that f(t) is piecewise continuous on t0 and has an exponential
ordera:Then the Laplace transform
F(s) =Z1
0f(t)e stdt
exists as long as s>a: Note that the two conditions above are sucient, but
not necessary, for F(s) to exist.
21 ESSENTIALS OF THE LAPLACE TRANSFORM 181
In what follows, we will denote the class of all piecewise continuous functions
with an exponential order by PE:The next theorem shows that any linear
combination of functions in PEis also inPE:The same is true for the product
of two functions in PE:
Theorem 21.2
Suppose that f(t) andg(t) are two elements of PEwith
jf(t)jM1ea1t; tC1andjg(t)jM2ea1t; tC2:
(i) For any constants and the function f(t) + g(t) is also a member of
PE. Moreover
L[ f(t) + g(t)] = L[f(t)] + L[g(t)]:
(ii) The function h(t) =f(t)g(t) is an element of PE:
We next discuss the problem of how to determine the function f(t) ifF(s)
is given. That is, how do we invert the transform. The following result on
uniqueness provides a possible answer. This result establishes a one-to-one
correspondence between the set PEand its Laplace transforms. Alterna-
tively, the following theorem asserts that the Laplace transform of a member
inPEis unique.
Theorem 21.3
Letf(t) andg(t) be two elements in PEwith Laplace transforms F(s) and
G(s) such that F(s) =G(s) for somes > a: Thenf(t) =g(t) for allt0
where both functions are continuous.
With the above theorem, we can now ocially de ne the inverse Laplace
transform as follows: For a function f2PE whose Laplace transform is F;
we callftheinverse Laplace transform ofFand writef=L 1[F(s)]:
Symbolically
f(t) =L 1[F(s)]()F(s) =L[f(t)]:
Example 21.3
FindL 1 1
s 1
; s> 1:
182 THE LAPLACE TRANSFORM SOLUTIONS FOR PDES
Solution.
From Example 23.1(a), we have that L[eat] =1
s a; s>a: In particular, for
a= 1 we nd thatL[et] =1
s 1; s> 1:Hence,L 1 1
s 1
=et; t0:
The above theorem states that if f(t) is continuous and has a Laplace trans-
formF(s), then there is no other function that has the same Laplace trans-
form. To ndL 1[F(s)], we can inspect tables of Laplace transforms of
known functions to nd a particular f(t) that yields the given F(s):
When the function f(t) is not continuous, the uniqueness of the inverse
Laplace transform is not assured. The following example addresses the
uniqueness issue.
Example 21.4
Consider the two functions f(t) =H(t)H(3 t) andg(t) =H(t) H(t 3);
whereHis the Heaviside function de ned by
H(t) =1; t0
0; t< 0
(a) Are the two functions identical?
(b) Show thatL[f(t)] =L[g(t):
Solution.
(a) We have
f(t) =1;0t3
0; t> 3
and
g(t) =1;0t<3
0; t3
Sincef(3) = 1 and g(3) = 0; fandgare not identical.
(b) We have
L[f(t)] =L[g(t)] =Z3
0e stdt=1 e 3s
s;s> 0:
Thus, both functions f(t) andg(t) have the same Laplace transform even
though they are not identical. However, they are equal on the interval(s)
where they are both continuous
The inverse Laplace transform possesses a linear property as indicated in
the following result.
21 ESSENTIALS OF THE LAPLACE TRANSFORM 183
Theorem 21.4
Given two Laplace transforms F(s) andG(s) then
L 1[aF(s) +bG(s)] =aL 1[F(s)] +bL 1[G(s)]
for any constants aandb:
Convolution integrals are useful when nding the inverse Laplace transform
of products. They are de ned as follows: The convolution of two scalar
piecewise continuous functions f(t) andg(t) de ned for t0 is the integral
(fg)(t) =Zt
0f(t s)g(s)ds:
Example 21.5
Findfgwheref(t) =e tandg(t) = sint:
Solution.
Using integration by parts twice we arrive at
(fg)(t) =Rt
0e (t s)sinsds
=1
2
e (t s)(sins coss)t
0
=e t
2+1
2(sint cost)
Next, we state several properties of convolution product, which resemble
those of ordinary product.
Theorem 21.5
Letf(t);g(t), andk(t) be three piecewise continuous scalar functions de ned
fort0 andc1andc2are arbitrary constants. Then
(i)fg=gf(Commutative Law)
(ii) (fg)k=f(gk) (Associative Law)
(iii)f(c1g+c2k) =c1fg+c2fk(Distributive Law)
Example 21.6
Express the solution to the initial value problem y0+ y=g(t); y(0) =y0
in terms of a convolution integral.
184 THE LAPLACE TRANSFORM SOLUTIONS FOR PDES
Solution.
Solving this initial value problem by the method of integrating factor we nd
y(t) =e ty0+Zt
0e (t s)g(s)ds=e ty0+ (e tg)(t)
The following theorem, known as the Convolution Theorem, provides a way
for nding the Laplace transform of a convolution integral and also nding
the inverse Laplace transform of a product.
Theorem 21.6
Iff(t) andg(t) are elements inPEthen
L[(fg)(t)] =L[f(t)]L[g(t)] =F(s)G(s):
Thus, (fg)(t) =L 1[F(s)G(s)]:0
Example 21.7
Use the convolution theorem to nd the inverse Laplace transform of
P(s) =1
(s2+a2)2:
Solution.
Note that
P(s) =1
s2+a21
s2+a2
:
So, in this case we have, F(s) =G(s) =1
s2+a2so thatf(t) =g(t) =1
asin (at):
Thus,
(fg)(t) =1
a2Zt
0sin (at as) sin (as)ds=1
2a3(sin (at) atcos (at))
The next example provides a solution method for solving ordinary di erential
equations.
Example 21.8
Solve the initial value problem
4y00+y=g(t); y(0) = 3; y0(0) = 7
21 ESSENTIALS OF THE LAPLACE TRANSFORM 185
Solution.
Take the Laplace transform of all the terms and plug in the initial conditions
to obtain
4(s2Y(s) 3s+ 7) +Y(s) =G(s)
or
(4s2+ 1)Y(s) 12s+ 28 =G(s):
Solving for Y(s) we nd
Y(s) =12s 28
4
s2+1
4+G(s)
4
s2+1
4
=3s
s2+
(1
22 7 1
22
s2+ 1
22+1
4G(s) 1
22
s2+ 1
22
Hence,
y(t) = 3 cost
2
7 sint
2
+1
2Zt
0sins
2
g(t s)ds:
So, once we decide on a g(t) all we need to do is to evaluate the integral and
we'll have the solution
We conclude this section with the following table of Laplace transform pairs.
186 THE LAPLACE TRANSFORM SOLUTIONS FOR PDES
f(t) F(s)
H(t) =1; t0
0; t< 01
s; s> 0
tn; n= 1;2;n!
sn+1; s> 0
e t 1
s ; s>
sin (!t)!
s2+!2; s> 0
cos (!t)s
s2+!2; s> 0
sinh (!t)!
s2 !2; s>j!j
cosh (!t)s
s2 !2; s>j!j
e tf(t); withjf(t)jMeatF(s ); s> +a
e ttn; n= 1;2;n!
(s )n+1; s>
e tsin (!t)!
(s )2+!2; s>
e tcos (!t)s
(s )2+!2; s>
f(t )H(t ); 0 e sF(s); s>a
withjf(t)jMeat
H(t ); 0e s
s; s> 0
tf(t) - F0(s)
t
2!sin!ts
(s2+!2)2; s> 0
1
2!3[sin!t !tcos!t]1
(s2+!2)2; s> 0
f0(t); with f (t)continuous sF (s) f(0)
andjf0(t)jMeats>maxfa;0g+ 1
f00(t); with f0(t)continuous s2F(s) sf(0) f0(0)
andjf00(t)jMeats>maxfa;0g+ 1
f(n)(t); with f(n 1)(t)continuous snF(s) sn 1f(0)
andjf(n)(t)jMeat-sf(n 2)(0) f(n 1)(0)
s>maxfa;0g+ 1
Rt
0f(u)du; withjf(t)jMeatF(s)
s; s> maxfa;0g+ 1
TableL
21 ESSENTIALS OF THE LAPLACE TRANSFORM 187
Practice Problems
Exercise 21.1
Determine whether the integralR1
01
1+t2dtconverges. If the integral con-
verges, give its value.
Exercise 21.2
Determine whether the integralR1
0t
1+t2dtconverges. If the integral con-
verges, give its value.
Exercise 21.3
Determine whether the integralR1
0e tcos (e t)dtconverges. If the integral
converges, give its value.
Exercise 21.4
Using the de nition, nd L[e3t];if it exists. If the Laplace transform exists
then nd the domain of F(s):
Exercise 21.5
Using the de nition, nd L[t 5];if it exists. If the Laplace transform exists
then nd the domain of F(s):
Exercise 21.6
Using the de nition, nd L[e(t 1)2];if it exists. If the Laplace transform
exists then nd the domain of F(s):
Exercise 21.7
Using the de nition, nd L[(t 2)2];if it exists. If the Laplace transform
exists then nd the domain of F(s):
Exercise 21.8
Using the de nition, nd L[f(t)];if it exists. If the Laplace transform exists
then nd the domain of F(s):
f(t) =0;0t<1
t 1; t1
188 THE LAPLACE TRANSFORM SOLUTIONS FOR PDES
Exercise 21.9
Using the de nition, nd L[f(t)];if it exists. If the Laplace transform exists
then nd the domain of F(s):
f(t) =8
<
:0;0t<1
t 1;1t<2
0; t2:
Exercise 21.10
Letnbe a positive integer. Using integration by parts establish the reduction
formula Z
tne stdt= tne st
s+n
sZ
tn 1e stdt; s> 0:
Exercise 21.11
Fors>0 andna positive integer evaluate the limits
(a) limt!0tne st(b) limt!1tne st
Exercise 21.12
Use the linearity property of Laplace transform to nd L[5e 7t+t+ 2e2t]:
Find the domain of F(s):
Exercise 21.13
FindL 1 3
s 2
:
Exercise 21.14
FindL 1
2
s2+1
s+1
:
Exercise 21.15
FindL 1 2
s+2+2
s 2
:
Exercise 21.16
Use TableLto ndL[2et+ 5]:
Exercise 21.17
Use TableLto ndL[e3t 3H(t 1)]:
Exercise 21.18
Use TableLto ndL[sin2!t]:
21 ESSENTIALS OF THE LAPLACE TRANSFORM 189
Exercise 21.19
Use TableLto ndL[sin 3tcos 3t]:
Exercise 21.20
Use TableLto ndL[e2tcos 3t]:
Exercise 21.21
Use TableLto ndL[e4t(t2+ 3t+ 5)]:
Exercise 21.22
Use TableLto ndL 1[10
s2+25+4
s 3]:
Exercise 21.23
Use TableLto ndL 1[5
(s 3)4]:
190 THE LAPLACE TRANSFORM SOLUTIONS FOR PDES
Sample Exam Questions
Exercise 21.24
Use TableLto ndL 1[e 2s
s 9]:
Exercise 21.25
Using the partial fraction decomposition nd L 1h
12
(s 3)(s+1)i
:
Exercise 21.26
Using the partial fraction decomposition nd L 1h
24e 5s
s2 9i
:
Exercise 21.27
Use Laplace transform technique to solve the initial value problem
y0+ 4y=g(t); y(0) = 2
where
g(t) =8
<
:0;0t<1
12;1t<3
0; t3
Exercise 21.28
Use Laplace transform technique to solve the initial value problem
y00 4y=e3t; y(0) = 0; y0(0) = 0:
Exercise 21.29
Consider the functions f(t) =etandg(t) =e 2t; t0:Computefgin
two di erent ways.
(a) By directly evaluating the integral.
(b) By computing L 1[F(s)G(s)] whereF(s) =L[f(t)] andG(s) =L[g(t)]:
Exercise 21.30
Consider the functions f(t) = sintandg(t) = cost; t0:Computefgin
two di erent ways.
(a) By directly evaluating the integral.
(b) By computing L 1[F(s)G(s)] whereF(s) =L[f(t)] andG(s) =L[g(t)]:
Exercise 21.31
Computettt.
21 ESSENTIALS OF THE LAPLACE TRANSFORM 191
Exercise 21.32
ComputeH(t)e te 2t.
Exercise 21.33
Computete tet.
192 THE LAPLACE TRANSFORM SOLUTIONS FOR PDES
22 Solving PDEs Using Laplace Transform
The same idea for solving linear ODEs using Laplace transform can be ex-
ploited when solving PDEs for functions in two variables u=u(x;t):The
transformation will be done with respect to the time variable t0;the spa-
tial variable xwill be treated as a parameter una ected by this transform.
In particular we de ne the Laplace transform of u(x;t) by the formula
L(u(x;t)) =U(x;s) =Z1
0u(x; )e s d :
The time derivatives are transformed in the same way as in the case of
functions in one variable, that is, for example
L(ut)(x;t) =sU(x;s) u(x;0)
and
L(utt)(x;s) =s2U(x;s) su(x;0) ut(x;0):
The spatial derivatives remain unchanged, for example,
Lux(x;t) =Z1
0ux(x; )e s d =@
@xZ1
0u(x; )e s d =Ux(x;s):
Likewise, we have
Luxx(x;t) =Uxx(x;s):
Thus, applying the Laplace transform to a PDE in two variables xandtwe
obtain an ODE in the variable xand with the parameter s:
Example 22.1
Letu(x;t) be the concentration of a chemical contaminant dissolved in a
liquid on a half-in nte domain x>0:Let us assume that at time t= 0 the
concentration is 0 and on the boundary x= 0;constant unit concentration of
the contaminant is kept for t>0:The behaviour of this problem is described
by the following mathematical model
8
>><
>>:ut uxx= 0; x> 0; t> 0
u(x;0) = 0;
u(0;t) = 1;
ju(x;t)j<1;8x>0;t> 0:
Findu(x;t):
22 SOLVING PDES USING LAPLACE TRANSFORM 193
Solution.
Applying Laplace transform to both sides of the equation we obtain
sU(x;s) u(x;0) Uxx(x;s) = 0
or
Uxx(x;s) sU(x;s) = 0:
This is a second order linear ODE in the variable xand positive parameter
s:Its general solution is
U(x;s) =A(s)e psx+B(s)epsx:
SinceU(x;s) is bounded in the variable x;we must have B(s) = 0 and in
this case we obtain
U(x;s) =A(s)e psx:
Next, we apply Laplace transform to the boundary condition obtaining
U(0;s) =L(1) =1
s:
This leads to A(s) =1
sand the transformed solution becomes
U(x;s) =1
se psx:
Thus,
u(x;t) =L 11
se psx
:
One can use a software package to nd the expression for L 1 1
se psx
Example 22.2
Solve the following initial boundary value problem
8
>><
>>:ut uxx= 0; x> 0; t> 0
u(x;0) = 0;
u(0;t) =f(t);
ju(x;t)j<1;8x>0;t> 0:
194 THE LAPLACE TRANSFORM SOLUTIONS FOR PDES
Solution.
Following the argument of the previous example we nd
U(x;s) =F(s)e psx; F(s) =Lf(t):
Thus, using Theorem 21.6 we can write
u(x;t) =L 1
F(s)e psx
=fL 1(e psx):
It can be shown that
L 1(e psx) =xp
4t3e x2
4t:
Hence,
u(x;t) =Zt
0xp
4(t s)3e x2
4(t s)f(s)ds
Example 22.3
Solve the wave equation
8
>><
>>:utt c2uxx= 0; x> 0; t> 0
u(x;0) =ut(x;0) = 0;
u(0;t) =f(t);
ju(x;t)j<1;8x>0;t> 0:
Solution.
Applying Laplace transform to both sides of the equation we obtain
s2U(x;s) su(x;0) ut(x;0) c2Uxx(x;s) = 0
or
c2Uxx(x;s) s2U(x;s) = 0:
This is a second order linear ODE in the variable xand positive parameter
s:Its general solution is
U(x;s) =A(s)e s
cx+B(s)es
cx:
SinceU(x;s) is bounded in the variable x;we must have B(s) = 0 and in
this case we obtain
U(x;s) =A(s)e s
cx:
22 SOLVING PDES USING LAPLACE TRANSFORM 195
Next, we apply Laplace transform to the boundary condition obtaining
U(0;s) =L(f(t)) =F(s):
This leads to A(s) =F(s) and the transformed solution becomes
U(x;s) =F(s)e s
cx:
Thus,
u(x;t) =L 1
F(s)e x
cs
=H
t x
c
f
t x
c
Remark 22.1
Laplace transforms are useful in solving parabolic and some hyperbolic PDEs.
They are not in general useful in solving elliptic PDEs.
196 THE LAPLACE TRANSFORM SOLUTIONS FOR PDES
Practice Problems
Exercise 22.1
Solve by Laplace transform
8
<
:ut+ux= 0; x> 0; t> 0
u(x;0) = sinx;
u(0;t) = 0
Hint: Method of integrating factor of ODEs.
Exercise 22.2
Solve by Laplace transform
8
<
:ut+ux= u ; x> 0; t> 0
u(x;0) = sinx;
u(0;t) = 0
Exercise 22.3
Solve
ut= 4uxx
u(0;t) =u(1;t) = 0
u(x;0) = 2 sinx+ 3 sin 2x:
Hint: A particular solution of a second order ODE must be found using the
method of variation of parameters.
Exercise 22.4
Solve by Laplace transform
8
<
:ut ux=u ; x> 0; t> 0
u(x;0) =e 5x;
ju(x;t)j<1
Exercise 22.5
Solve by Laplace transform
8
<
:ut+ux=t ; x> 0; t> 0
u(x;0) = 0;
u(0;t) =t2
22 SOLVING PDES USING LAPLACE TRANSFORM 197
Exercise 22.6
Solve by Laplace transform
8
<
:xut+ux= 0; x> 0; t> 0
u(x;0) = 0;
u(0;t) =t
Exercise 22.7
Solve by Laplace transform
8
>><
>>:utt c2uxx= 0; x> 0; t> 0
u(x;0) =ut(x;0) = 0;
u(0;t) = sinx;
ju(x;t)j<1
Exercise 22.8
Solve by Laplace transform
utt 9uxx= 0;0x; t> 0
u(0;t) =u(;t) = 0;
ut(x;0) = 0; u(x;0) = 2 sinx:
Exercise 22.9
Solve by Laplace transform
8
<
:uxy= 1; x> 0; y> 0
u(x;0) = 1;
u(0;y) =y+ 1:
Exercise 22.10
Solve by Laplace transform
8
>><
>>:utt=c2uxx; x> 0; t> 0
u(x;0) =ut(x;0) = 0;
ux(0;t) =f(t);
ju(x;t)j<1:
198 THE LAPLACE TRANSFORM SOLUTIONS FOR PDES
Sample Exam Questions
Exercise 22.11
Solve by Laplace transform
8
<
:ut+ux=u ; x> 0; t> 0
u(x;0) = sinx;
u(0;t) = 0
Exercise 22.12
Solve by Laplace transform
8
>><
>>:ut c2uxx= 0; x> 0; t> 0
u(x;0) =T;
u(0;t) = 0;
ju(x;t)j<1
Exercise 22.13
Solve by Laplace transform
ut 3uxx= 0;0x2; t> 0
u(0;t) =u(2;t) = 0;
u(x;0) = 5 sin (x)
Exercise 22.14
Solve by Laplace transform
ut 4uxx= 0;0x; t> 0
ux(0;t) =u(;t) = 0;
u(x;0) = 40 cosx
2
Exercise 22.15
Solve by Laplace transform
utt 4uxx= 0;0x2; t> 0
u(0;t) =u(2;t) = 0;
ut(x;0) = 0; u(x;0) = 3 sinx:
The Fourier Transform
Solutions for PDEs
In the previous chapter we discussed one class of integral transform meth-
ods, the Laplace transfom. In this chapter, we consider a second fundamental
class of integral transform methods, the so-called Fourier transform .
Fourier series are designed to solve boundary value problems on bounded
intervals. The extension of Fourier methods to the entire real line leads nat-
urally to the Fourier transform, an extremely powerful mathematical tool for
the analysis of non-periodic functions. The Fourier transform is of fundamen-
tal importance in a broad range of applications, including both ordinary and
partial di erential equations, quantum mechanics, signal processing, control
theory, and probability, to name but a few.
23 Complex Version of Fourier Series
We have seen in Section 15 that a 2 L periodic function f:R!Rthat is
piecewise smooth on [ L;L] can be expanded in a Fourier series
f(x) =a0
2+1X
n=1
ancosn
Lx
+bnsinn
Lx
at all points of continuity of f:In the context of Fourier analysis, this is
referred to as the real form of the Fourier series. It is often convenient to
recast this series in complex form by means of Euler formula
eix= cosx+isinx:
It follows from this formula that
199
200 THE FOURIER TRANSFORM SOLUTIONS FOR PDES
eix+e ix= 2 cosxandeix e ix= 2isinx
or
cosx=eix+e ix
2and sinx=eix e ix
2i:
Hence the Fourier expansion of fcan be rewritten as
f(x) =a0
2+1X
n=1"
an
einx
L+e inx
L
2!
+bn
einx
L e inx
L
2i!#
f(x) =1X
n= 1cneinx
L (23.1)
wherec0=a0
2and forn2Nwe have
cn=an ibn
2
c n=an+ibn
2:
It follows that if n2Nthen
an=cn+c nandbn=i(cn c n): (23.2)
That is,anandbncan be easily found once we have formulas for cn:In order
to nd these formulas, we need to evaluate the following integral
ZL
Leinx
Le imx
Ldx=ZL
Lei(n m)x
Ldx
=L
i(n m)ei(n m)x
LiL
L
= iL
(n m)[cos [(n m)] +isin [(n m)]
cos [ (n m)] isin [ (n m)]]
=0
23 COMPLEX VERSION OF FOURIER SERIES 201
ifn6=m:Ifn=mthen
ZL
Leinx
Le inx
Ldx= 2L:
Now, if we multiply (23.1) by e inx
Land integrate from LtoLand apply
the last result we ndZL
Lf(x)e inx
Ldx= 2Lcn
which yields the formula for coecients of the complex form of the Fourier
series:
cn=1
2LZL
Lf(x)e inx
Ldx; n = 0;1;2;:
Example 23.1
Find the complex Fourier coecients of the function
f(x) =x; x
extended to be periodic of period 2 :
Solution.
Using integration by parts and the fact that ei=e i= 1 we nd
cn=1
2Z
xe inxdx
=1
2ix
n
e inx
Z
i
n
e inxdx
=1
2i
n
e in+i
n
ein
+1
21
n2e in 1
n2ein
=1
2h
2i
n( 1)ni
+1
2(0) =( 1)ni
n
Remark 23.1
It is often the case that the complex form of the Fourier series is far simpler
to calculate than the real form. One can then use (23.2) to nd the real form
of the Fourier series. For example, the Fourier coecients of the real form of
the previous function are given by
an= (cn+c n) = 0 andbn=i(cn c n) =2
n( 1)n+1; n2N
202 THE FOURIER TRANSFORM SOLUTIONS FOR PDES
Practice Problems
Exercise 23.1
Find the complex Fourier coecients of the function
f(x) =x; 1x1
extended to be periodic of period 2 :
Exercise 23.2
Let
f(x) =8
<
:0 <x<
2
1
2<x<
2
0
2<x<
be 2 periodic. Find its complex series representation.
Exercise 23.3
Find the complex Fourier series of the 2 periodic function f(x) =eaxover
the interval ( ;):
Exercise 23.4
Find the complex Fourier series of the 2 periodic function f(x) = sinx
over the interval ( ;):
Exercise 23.5
Find the complex Fourier series of the 2 periodic function de ned
f(x) =1 0<x<T
0T <x< 2
Exercise 23.6
Letf(x) =x2; <x<; be 2 periodic.
(a) Calculate the complex Fourier series representation of f:
(b) Using the complex Fourier series found in (a), recover the real Fourier
series representation of f:
Exercise 23.7
Letf(x) = sinnx; 1
2<x<1
2;be of period 1.
(a) Calculate the coecients an;bnandcn:
(b) Find the complex Fourier series representation of f:
23 COMPLEX VERSION OF FOURIER SERIES 203
Exercise 23.8
Letf(x) = 2 x; 2<x< 2;be of period 2.
(a) Calculate the coecients an;bnandcn:
(b) Find the complex Fourier series representation of f:
Exercise 23.9
Suppose that the coecients cnof the complex Fourier series are given by
cn=2
inifjnjis odd
0 ifjnjis even.
Findan; n= 0;1;2;andbn; n= 1;2;:
Exercise 23.10
Recall that any complex number zcan be written as z=Re(z) +iIm(z)
whereRe(z) is called the real part of zandIm(z) is called the imaginary
part . The complex conjugate ofzis the complex number z=Re(z)
iIm(z):Using these de nitions show that an= 2Re(cn) andbn= 2Im(cn):
204 THE FOURIER TRANSFORM SOLUTIONS FOR PDES
Sample Exam Questions
Exercise 23.11
Suppose that
cn=i
2n[e inT 1] ifn6= 0
T
2ifn= 0:
Findanandbn:
Exercise 23.12
Find the complex Fourier series of the function f(x) =exon [ 2;2]:
Exercise 23.13
Consider the wave form
(a) Writef(x) explicitly. What is the period of f:
(b) Determine a0andanforn2N:
(c) Determine bnforn2N:
(d) Determine c0andcnforn2N:
Exercise 23.14
Ifzis a complex number we de ne sin z=1
2(eiz e iz):Find the complex
form of the Fourier series for sin 3 xwithout evaluating any integrals.
Exercise 23.15
Findcnfor the 2 periodic function
f(x) =1 ifsxs+h
0 elsewhere in [ ;]
24 THE ONE DIMENSIONAL FOURIER TRANSFORM 205
24 The One Dimensional Fourier Transform
One of the problems with the theory of Fourier series discussed so far is that
it applies only to periodic functions. There are many times when one would
like to divide a function which is not periodic into a superposition of sines
and cosines. The Fourier transform is the tool often used for this purpose.
Like the Laplace transform, the Fourier transform is often an e ective tool
in nding explicit solutions to di erential equations.
To start with, let f:R!Rbe a piecewise continuous function that vanishes
outside an interval of the form [ L;L ]:This function can be extended to
a periodic function, still denoted by f, of period 2 L:From the previous
section we can nd the complex Fourier series of fto be
f(x) =1X
n= 1cneinx
L (24.1)
where
cn=1
2LZL
Lf(x)e inx
Ldx:
Let2R:Multiply both sides of (24.1) by e ixand then integrate both sides
from LtoL:Assuming integration and summation can be interchanged
we ndZL
Lf(x)e ixdx=1X
n= 1cnZL
Le ixeinx
Ldx:
It can be shown that the RHS converges, say to ^f();asL!1:Hence, we
nd
^f() =Z1
1f(x)e ixdx: (24.2)
Thus, for a piecewise continuous function f;we de ne the Fourier trans-
form offto be the function ^fgiven by (24.2). We will use the notation
F[f(x)] = ^f():
Now, letting =n
Lin (24.2) we nd
^fn
L
=ZL
Lf(x)e inx
Ldx= 2Lcn:
206 THE FOURIER TRANSFORM SOLUTIONS FOR PDES
Hence, (24.1) can be written in the form
f(x) =1
2L1X
n= 1^fn
L
einx
L:
In the limit as L!1;it can be shown that this last sum approaches an
improper integral, and our formula becomes
F 1[^f()] =f(x) =1
2Z1
1^f()eixd (24.3)
Equation (24.3) is called the Fourier inversion formula . If we make use of
Euler's formula, we can write the Fourier inversion formula in terms of sines
and cosines,
f(x) =1
2Z1
1^f() cosxd +i
2Z1
1^f() sinxd
a superposition of sines and cosines of various frequencies.
Equations (24.2) and (24.3) allow one to pass back and forth between a given
function and its representation as a superposition of oscillations of various
frequencies.
Example 24.1
Find the Fourier transform of the function f(x) de ned by
f(x) =e axifx0
0 ifx<0
for somea>0:
Solution.
We have
^f() =Z1
1f(x)e ixdx=Z1
0e axe ixdx
=Z1
0e ax ixdx=e x(a+i)
(a+i) 1
0
=1
a+i
The following theorem lists the basic properties of Fourier transform
24 THE ONE DIMENSIONAL FOURIER TRANSFORM 207
Theorem 24.1
Letf;g; be piecewise continuous functions. Then we have the following
properties:
(1)Linearity:F[ f(x) + g(x)] = F[f(x)] + F[g(x)];where and are
arbitrary numbers.
(2)Shifting :F[f(x )] =e i F[f(x)]:
(3)Scaling:F[f x
] = F[f( x)]:
(4)Continuity: IfR1
1jf(x)jdx<1then ^fis continuous in .
(5)Di erentiation: F[f(n)(x)] = (i)nF[f(x)):
(6)Integration:FRx
0f(s)ds
= 1
iF[f(x)]:
(7)Parseval's Relation:R1
1jf(x)j2dx=1
2R1
1j^f()j2d:
(8)Duality:F[F[f(x)]] = 2f( x):
(9)Multiplication by xn:F[xnf(x)] =in^f(n)():
(10)Gaussians:F[e x2] =p
e 2
4 :
(11)Product:F[(f(x)g(x)] =1
2F[f(x)]F[g(x)]:
(12)Convolution:F[(fg)(x)] =F[f(x)]F[g(x)]:
Example 24.2
Determine the Fourier transform of the Gaussian u(x) =e x2; > 0:
Solution.
We have
^u() =Z1
1e x2e ixdx:
If we di erentiate this relation with respect to the variable and then inte-
grate by parts we obtain
^u0() = iZ1
1xe x2e ixdx
=i
2 Z1
1d
dx(e x2)e ixdx
=i
2 Z1
1(e x2)e ixdx=
2 ^u()
Thus we have arrived at the ODE ^ u0() =
2 ^u() whose general solution
has the form
^u() =Ce 2
4
208 THE FOURIER TRANSFORM SOLUTIONS FOR PDES
Since
^u(0) =Z1
1e x2dx=r
=C
we nd
^u() =r
e 2
4
Example 24.3
Prove
F[f( x)] = ^f( ):
Solution.
Using a change of variables we nd
F[f( x)] =Z1
1f( x)e ixdx=Z1
1f(x)eixdx=^f( )
Example 24.4
Prove
F[F[f(x)]] = 2f( x):
Solution.
We have
f(x) =1
2Z1
1^f()eixd
Thus,
2f( x) =Z1
1^f()e ixd=F[^f()] =F[F[f(x)]]
The following theorem lists the properties of inverse Fourier transform
Theorem 24.2
Letfandgbe piecewise continuous functions.
(1')Linearity:F 1[ ^f() + ^g()] = F 1[^f()] + F 1[^g()]:
(2')Derivatives:F 1[^f(n)()] = ( ix)nf(x):
(3')Multiplication by n:F 1[n^f()] = ( i)nf(n)(x):
(4')Multiplication by e i :F 1[e i ^f()] =f(x ):
(5')Gaussians:F 1[e 2] =1p
4 e x2
4 :
(6')Product:F 1[^f()^g()] =f(x)g(x):
(7')Convolution:F 1[^f^g()] = 2(fg)(x):
24 THE ONE DIMENSIONAL FOURIER TRANSFORM 209
Remark 24.1
It is important to mention that there exists no established convention how
to de ne the Fourier transform. In the literature, we can meet an equivalent
de nition of (24.2) with the constant1p
2or1
2in front of the integral.
There also exist de nitions with positive sign in the exponent. The reader
should keep this fact in mind while working with various sources or using the
transformation tables.
210 THE FOURIER TRANSFORM SOLUTIONS FOR PDES
Practice Problems
Exercise 24.1
Find the Fourier transform of the function
f(x) =1 if 1x1
0 otherwise.
Exercise 24.2
Obtain the transformed problem when applying the Fourier transform with
respect to the spatial variable to the equation and initial condition
ut+cux= 0
u(x;0) =f(x):
Exercise 24.3
Obtain the transformed problem when applying the Fourier transform with
respect to the spatial variable to the equation and both initial conditions
utt=c2uxx; x2R; t> 0
u(x;0) =f(x)
ut(x;0) =g(x):
Exercise 24.4
Obtain the transformed problem when applying the Fourier transform with
respect to the spatial variable to the equation and both initial conditions
u=uxx+uyy= 0; x2R;0<y<L
u(x;0) = 0
u(x;L) =1 if a<x<a
0 otherwise
Exercise 24.5
Find the Fourier transform of f(x) =e jxj ;where >0:
24 THE ONE DIMENSIONAL FOURIER TRANSFORM 211
Exercise 24.6
Prove that
F[e xH(x)] =1
1 +i
where
H(x) =1 ifx0
0 otherwise.
Exercise 24.7
Prove that
F1
1 +ix
= 2eH( ):
Exercise 24.8
Prove
F[f(x )] =e i ^f():
Exercise 24.9
Prove
F[ei xf(x)] = ^f( ):
Exercise 24.10
Prove the following
F[cos ( x)f(x)] =1
2[^f(+ ) +^f( )]
F[sin ( x)f(x)] =i
2[^f(+ ) ^f( )]
212 THE FOURIER TRANSFORM SOLUTIONS FOR PDES
Sample Exam Questions
Exercise 24.11
Prove
F[f0(x)] = (i)^f():
Exercise 24.12
Find the Fourier transform of f(x) = 1 jxjfor 1x1 and 0 otherwise.
Exercise 24.13
Find, using the de nition, the Fourier transform of
f(x) =8
<
: 1 a<x< 0
1 0<x<a
0 otherwise
Exercise 24.14
Find the inverse Fourier transform of ^f() =e 2
2:
Exercise 24.15
FindF 1
1
a+i
:
25 APPLICATIONS OF FOURIER TRANSFORMS TO PDES 213
25 Applications of Fourier Transforms to PDEs
Fourier transform is a useful tool for solving di erential equations. In this
section, we apply Fourier transforms in solving various PDE problems. Con-
trary to Laplace transform, which usually uses the time variable, the Fourier
transform is applied to the spatial variable on the whole real line.
The Fourier transform will be applied to the spatial variable xwhile the vari-
abletremains xed. The PDE in the two variables xandtpasses under the
Fourier transform to an ODE in the t variable. We solve this ODE to obtain
the transformed solution ^ uwhich can be converted to the original solution u
by means of the inverse Fourier transform. We illustrate these ideas in the
examples below.
First Order Transport Equation
Consider the initial value problem
ut+cux= 0
u(x;0) =f(x):
Let ^u(;t) be the Fourier transform of uinx:Performing the Fourier trans-
form on both the PDE and the initial condition, we reduce the PDE into an
ODE int@^u
@t+ic^u= 0
^u(;0) = ^f():
Solution of the ODE gives
^u(;t) =^f()e ict:
Thus,
u(x;t) =F 1[^u(;t)] =f(x ct)
which is exactly the same as we obtained by using the method of character-
istics.
Second Order Wave Equation
Consider the two dimensional wave equation
utt=c2uxx; x2R; t> 0
214 THE FOURIER TRANSFORM SOLUTIONS FOR PDES
u(x;0) =f(x)
ut(x;0) =g(x):
Again, by performing the Fourier transform of uinx;we reduce the PDE
problem into an ODE problem in the variable t:
@2^u
@t2= c22^u
^u(;0) = ^f()
^ut(;0) = ^g():
General solution to the ODE is
^u(;t) = ()e ict+ ()eict
where and are two arbitrary functions of :Performing the inverse
transformation and making use of the translation theorem, we get the general
solution
u(x;t) = (x ct) + (x+ct)
where ^ = and ^ = :But
() =1
2
^f() 1
ic^g()
() =1
2
^f() +1
ic^g()
:
By using the integration property, we nd the inverse transforms of and
(x) =1
2
f(x) +1
cZx
0g(s)ds
(x) =1
2
f(x) 1
cZx
0g(s)ds
:
Application of the translation property then yields directly the D'Alambert
solution
u(x;t) =1
2[f(x ct) +f(x+ct)] +1
2cZx+ct
x ctg(s)ds:
25 APPLICATIONS OF FOURIER TRANSFORMS TO PDES 215
Second Order Heat Equation
Next, we consider the heat equation
ut=kuxx; x2R; t> 0
u(x;0) =f(x):
Performing Fourier Transform in xfor the PDE and the initial condition, we
obtain@^u
@t= k2^u
^u(;0) = ^f():
Treatingas a parameter, we obtain the solution to the above ODE problem
^u(;t) =^f()e k2t:
Application of the convolution theorem yields
u(x;t) =f(x)F 1[e k2t]
=f(x)1p
4kte x2
4kt
=1p
4ktZ1
1f(s)e (x s)2
4ktds
Laplace's Equation in 2D
Consider the problem
u=uxx+uyy= 0; x2R;0<y<L
u(x;0) = 0
u(x;L) =1 if a<x<a
0 otherwise
Performing Fourier Transform in xfor the PDE we obtain the second order
PDE iny
^uyy=2^u:
The general solution is given by
^u(;y) =A() sinhy+B() coshy:
216 THE FOURIER TRANSFORM SOLUTIONS FOR PDES
Using the boundary condition ^ u(;0) = 0 we nd B() = 0:Using the second
boundary condition we nd
^u(;L) =Z1
1u(x;L)e ixdx
=Za
ae ixdx=Za
acosxdx
=2 sina
Hence,
A() sinhL=2 sina
and this implies
A() =2 sina
sinhL:
Thus,
^u(;y) =2 sina
sinhLsinhy:
Taking inverse Fourier transform we nd
u(x;y) =1
2Z1
12 sina
sinhLsinhyeixd:
Noting that the integrand is an even function in ;we can simplify a little to
obtain
u(x;y) =1
2Z1
12 sina
sinhLsinhycosxd
25 APPLICATIONS OF FOURIER TRANSFORMS TO PDES 217
Practice Problems
Exercise 25.1
Solve, by using Fourier transform
ut+cux= 0
u(x;0) =e x2
4:
Exercise 25.2
Solve, by using Fourier transform
ut=kuxx u; x2R
u(x;0) =e x2
:
Exercise 25.3
Solve the heat equation
ut=kuxx
subject to the initial condition
u(x;0) =1 ifx0
0 otherwise.
Exercise 25.4
Use Fourier transform to solve the heat equation
ut=uxx+u; 1<x<1< t> 0
u(x;0) =f(x):
Exercise 25.5
Prove that Z1
1e jjyeixd=2y
x2+y2:
Exercise 25.6
Solve the Laplace's equation in the half plane
uxx+uyy= 0; 1<x<1;0<y<1
subject to the boundary condition
u(x;0) =f(x);ju(x;y)j<1:
218 THE FOURIER TRANSFORM SOLUTIONS FOR PDES
Exercise 25.7
Use Fourier transform to nd the transformed equation of
utt+ ( + )ut+ u =c2uxx
where ; > 0:
Exercise 25.8
Solve the initial value problem
ut+ 3ux= 0
u(x;0) =e x
using the Fourier transform.
Exercise 25.9
Solve the initial value problem
ut=kuxx
u(x;0) =e x
using the Fourier transform.
25 APPLICATIONS OF FOURIER TRANSFORMS TO PDES 219
Sample Exam Questions
Exercise 25.10
Solve the initial value problem
ut=kuxx
u(x;0) =e x2
using the Fourier transform.
Exercise 25.11
Solve the initial value problem
ut+cux= 0
u(x;0) =x2
using the Fourier transform.
Exercise 25.12
Solve, by using Fourier transform
u= 0
uy(x;0) =f(x)
lim
x2+y2!1u(x;y) = 0:
220 THE FOURIER TRANSFORM SOLUTIONS FOR PDES
Answers and Solutions
Section 1
1.1
(a)y3cos (xy)
(b)ex2y(2y+ 4x2y2)
(c) 0
1.2
(a)fx(x;y) = 4x3; fy(x;y) =3py
(b)fx(x;y;z ) = 2xy+ 43; fy(x;y;z ) =x2 20yz3 28
1+16y2; fz(x;y;z ) =
30y2z2
(c)fs(s;t) =2t7
s 4
7s 3
7; ft(s;t) = 7t6ln (s2) 27
t4
(d)fx(x;y) =4
x2sin 4
x
ex2y 5y3+cos 4
x
ex2y 5y3(2xy); fy(x;y) = cos 4
x
ex2y 5y3(x2
15y2)
(e)fu(u;v) =9(u2+5v) 9u(2u)
(u2+5v)2 = 9u2+45v
(u2+5v)2; fv(u;v) = 45u
(u2+5v)2
(f)fx(x;y;z ) =siny
z2; fy(x;y;z ) =xcosy
z2; fz(x;y;z ) = 2xsiny
z3
(g)fx(x;y) =1
2
2x+5
5x 3y2
(x2+ln (5x 3y2)) 1
2; fy(x;y) = 3y
5x 3y2(x2+
ln (5x 3y2)) 1
2
1.33
1.4
@z
@s=t2est2sin (s2t) + 2stest2cos (s2t)
@z
@t= 2stest2sin (s2t) +s2est2cos (s2t)
1.5uis the depedent variable whereas xandyare the independent variables.
221
222 ANSWERS AND SOLUTIONS
1.6We have Za
af(x)dx=Z0
af(x)dx+Za
0f(x)dx:
By the change of variable u= xwe nd
Z0
af(u)du= Z0
af( u)du= Za
0f(u)du:
Hence, the result follows.
1.7We have Za
af(x)dx=Z0
af(x)dx+Za
0f(x)dx:
By the change of variable u= xwe nd
Z0
af(u)du= Z0
af( u)du=Za
0f(u)du:
Hence, the result follows.
1.8By the product rule of derivatives we have
(uv)0=u0v+uv0:
Integrate both sides to obtain
uv=Z
u0vdx+Z
uv0dx:
Now subtractR
u0vdxfrom both sides to obtain the desired result.
1.9
utt= sin x
sin t
uxx= sin x
sin t
:
1.10
utt= sin x
sinh t
uxx= sin x
sinh t
:
223
1.112supf sinh t
g
1.12 (a) We have supfjun(x;0) 1j:x2Rg=1
nsupfjsinnxj:x2Rg=1
n:
(b) We have supfjun(x;t) 1j:x2Rg=en2t
n
Section 2
2.1(a) For all 0x < 1 we have lim n!1fn(x) = limn!1xn= 0:Also,
limn!1fn(1) = 1:Hence, the sequence ffng1
n=1converges pointwise to f:
(b) Suppose the contrary. Let =1
2:Then there exists a positive integer N
such that for all nNwe have
jfn(x) f(x)j<1
2
for allx2[0;1]:In particular, we have
jfN(x) f(x)j<1
2
for allx2[0;1]:Choose (0:5)1
N<x< 1:ThenjfN(x) f(x)j=xN>0:5 =
which is a contradiction. Hence, the given sequence does not converge uni-
formly.
2.2For every real number x, we have
lim
n!1fn(x) = lim
n!1nx+x2
n2= lim
n!1x
n+ lim
n!1x2
n2= 0
Thus,ffng1
n=1converges pointwise to the zero function on R:
2.3For every real number x, we have
1pn+ 1fn(x)1pn+ 1:
Moreover,
lim
n!11pn+ 1= 0:
Applying the squeeze rule for sequences, we obtain
lim
n!1fn(x) = 0
224 ANSWERS AND SOLUTIONS
for allxinR:Thus,ffng1
n=1converges pointwise to the zero function on R:
2.4First of all, observe that fn(0) = 0 for every ninN:So the sequence
ffn(0)g1
n=1is constant and converges to zero. Now suppose 0 <x< 1 then
n2xn=n2enlnx:But lnx<0 when 0<x< 1;it follows that
limn!1fn(x) = 0 for 0 <x< 1
Finally,fn(1) =n2for alln:So,
lim
n!1fn(1) =1:
Therefore,ffng1
n=1is not pointwise convergent on [0 ;1]:
2.5For
2x<0 and 0<x
2we have
lim
n!1(cosx)n= 0:
Forx= 0 we have fn(0) = 1 for all ninN:Therefore,ffng1
n=1converges
pointwise to
f(x) =0 if
2x<0 and 0<x
2
1 if x= 0:
2.6(a) Let >0 be given. Let Nbe a positive integer such that N >1
:
Then fornN x xn
n x =jxjn
n<1
n1
N<:
Thus, the given sequence converges uniformly (and pointwise) to the function
f(x) =x:
(b) Since lim n!1f0
n(x) = 1 for all x2[0;1), the sequenceff0
ng1
n=1converges
pointwise to f0(x) = 1:However, the convergence is not uniform. To see
this, let=1
2and suppose that the convergence is uniform. Then there is a
positive integer Nsuch that for nNwe have
j1 xn 1 1j=jxjn 1<1
2:
In particular, if we let n=N+ 1 we must have xN<1
2for allx2[0;1):
Butx= 1
21
N2[0;1) andxN=1
2which contradicts xN<1
2:Hence, the
225
convergence is not uniform.
2.7(a) The pointwise limit is
f(x) =8
<
:0 if 0x<1
1
2ifx= 1
1 if 1<x2
(b) The convergence cannot be uniform because if it were fwould have to
be continuous.
2.8(a) Let>0 be given. Note that
jfn(x) 1
2j= 2 cosx sin2x
2(2n+ sin2x) 3
4n:
Since lim n!13
4n= 0 we can nd a positive integer Nsuch that if nN
then3
4n<:Thus, fornNand allx2Rwe have
jfn(x) 1
2j3
4n<:
This shows that fn!1
2uniformly on Rand also on [2 ;7]:
(b) We have
lim
n!1Z7
2fnxdx=Z7
2lim
n!1fnxdx=Z7
21
2dx=5
2:
2.9We have proved earlier that this sequence converges pointwise to the
discontinuous function
f(x) =0 if
2x<0 and 0<x
2
1 if x= 0
Therefore, uniform convergence cannot occur for this given sequence.
2.10 (a) Using the squeeze rule we nd
lim
n!1supfjfn(x)j: 2x5g= 0:
226 ANSWERS AND SOLUTIONS
Thus,ffng1
n=1converges uniformly to the zero function.
(b) We have
lim
n!1Z5
2fn(x)dx=Z5
20dx= 0:
Section 3.
3.1y=1
2(1 e t2):
3.2y(t) =3t 1
9+e 2t+Ce 3t:
3.3y(t) = 3 sint+3 cost
t+C
t:
3.4y(t) =1
13(3 sin (3t) + 2 cos (3t)) +Ce 2t:
3.5y(t) =Ce sint 3:
3.6 = 2:
3.7p(t) = 2 andg(t) = 2t+ 3:
3.8y0=y(0) = 1 andg(t) = 2et+ cost+ sint:
3.91.
3.10u(x;y) =f(bx ay)e c
a2+b2(ax+by):
3.11y(t) =tlnjtj+ 7t:
3.12 Sincep(t) =awe nd(t) =eat. Suppose rst that a=. Then
y0+ay=be at
and the corresponding general solution is
y(t) =bte at+Ce at
Thus,
limt!1y(t) = lim t!1(bt
eat+C
eat)
= limt!1b
aeat= 0
227
Now, suppose that a6=then
y(t) =b
a e t+Ce at
Thus,
lim
t!1y(t) = 0:
3.13y(t) = ( tet+et) 1:
3.14y(t) =t2
4 t
3+t2
2+1
12t2:
3.15y(t) =tSi(t) + (3 Si(1))t:
Section 4
4.1y(t) =
3
2et2+C1
3:
4.2y(t) =Cet2
2 2t:
4.3y(t) =Ct2+ 4:
4.4y(t) =2Ce4t
1+Ce4t:
4.5y(t) =p
5 4 cos (2t):
4.6y(t) = p
( 2 cost+ 4):
4.7y(t) =e1 t 1:
4.8y(t) =2p
4t2+1:
4.9y(t) = tan (t+) = cott:
4.10y(t) =3 e t2
3+e t2:
4.11 =1
2; y0=1
2andn= 3:
4.12u(x;y) =F(y)e 3x+G(x) whereF(y) =R
f(y)dy:
228 ANSWERS AND SOLUTIONS
4.13y2+ cosy+ cost+t2
2= 2:
4.14 3y2y0+ cosy+ 2t= 0; y(2) = 0:
4.15 The ODE is not separable.
Section 5
5.1y(t) = 2et+e3t:limt! 1y(t) = 0 and lim t!1y(t) =1:
5.2y(t) = 2p
2e( 2 p
2)t+2p
2e( 2+p
2)t:limt! 1y(t) = 1and limt!1y(t) =
0:
5.3y(t) = 2e p
2
2t:limt! 1y(t) = 1 and limt!1y(t) = 0:
5.4y00 y0 2y= 0:
5.5y(t) =et
3 1(1 t):
5.6y(t) =e 2t
5(t 1):
5.7y(0) = 2 and y0(0) = 2:
5.8y(t) =c1e3t+c2te3t:
5.9y(t) = 3e tcost+ 2e tsint:
5.10y(t) = e1
2(t+)(3 cost
2+ sint
2):
5.11y(t) =yh(t) +yp(t) =e1
2t(c1cosp
3
2t+c2sinp
3
2t) +6
73cos 3t 16
73sin 3t:
5.12y(t) =yh(t) +yp(t) =c1e( 2 p
6)t+c2e( 2+p
6)t t2 5
2t 9:
5.13y=Ax4+Bx4lnx:
5.14y=x 1(Acos (p
3 lnx) +Bsin (p
3 lnx)):
229
5.15 (a)n=n2; yn(x) = sinnx; n = 1;2;:
(b)n=
n 1
2
L2andyn= sin
L
n 1
2
x
; n= 1;2;3;:
(c)n=
n 1
22; yn(x) = cos
n 1
2
x; n = 1;2;:
5.16 We consider rst the cases (a) and (b). Multiply the equation by y0(x)
and integrate in xfrom 0 toL:
ZL
0(ky0(x))0y(x)dx+ZL
0y2(x)dx= 0:
Use integration by parts in the rst integral
[ky0(x)y(x)]L
0 ZL
0k(y0(x))2dx+ZL
0y2(x)dx= 0:
The boundary term vanishes because of the boundary conditions. We solve
the above equation for and obtain
=RL
0k(y0(x))2dx
RL
0y2(x)dx0:
For the case (c), we repeat the above argument but by integrating from L
toL:
5.17y(t) = 2et
2:
5.18y(t) =c1+c2et 1
10cos (2t) +1
5sin (2t) + 5tet
5.19y(t) =17
15et+1
6e 2t 1
2t 1
4 3
20sin 2t 1
20cos 2t:
Section 6
6.1(a) ODE (b) PDE (c) ODE.
6.2uss= 0:
6.3uss+utt= 0:
6.4(a) Order 3, nonlinear (b) Order 1, linear, homogeneous (c) Order 2,
230 ANSWERS AND SOLUTIONS
linear, nonhomogeneous.
6.5(a) Linear, homogeneous, order 3.
(b) Linear, nonhomogeneous, order 3. The inhomogeneity is siny:
(c) Nonlinear, order 2. The nonlinear term is uux:
(d) Nonlinear, order 3. The nonlinear terms are uxuxxyanduuy:
(e) Linear, nonhomogeneous, order 2. The inhomogeneity is f(x;y;t ):
6.6(a) Linear. (b) Linear. (c) Nonlinear. (d) Nonlinear.
6.7(a) PDE, linear, second order, homogeneous.
(b) PDE, linear, second order, homogeneous.
(c) PDE, nonlinear, fourth order.
(d) ODE, linear, second order, nonhomogeneous.
(e) PDE, linear, second order, nonhomogeneous.
(f) PDE, quasilinear, second order.
6.8A(x;y;z )uxx+B(x;y;z )uxy+C(x;y;z )uyy+E(x;y;z )uxz+F(x;y;z )uyz+
G(x;y;z )uzz+H(x;y;z )ux+I(x;y;z )uy+J(x;y;z )uz+K(x;y;z )u=L(x;y;z ):
6.9(a) Order 3, linear, homogeneous.
(b) Order 1, nonlinear.
(c) Order 4, linear, nonhomogeneous
(d) Order 2, nonlinear.
(e) Order 2, linear, homogeneous.
6.10uww= 0:
6.11uvw= 0:
6.12uvw= 0:
6.13us= 0:
6.14us=1
2:
6.15uw=u:
231
Section 7
7.1a=b= 0:
7.2Substituting into the di erential equation we nd
tX00T XT0= 0
or
X00
X=T0
tT:
The LHS is a function of xonly whereas the RHS is a function of tonly.
This is true only when both sides are constant. That is, there is such that
X00
X=T0
tT=
and this leads to the two ODEs X00=XandT0=tT:
7.3We havexux+ (x+ 1)yuy=x
y(ex+xex) + (x+ 1)y
xex
y2
= 0 and
u(1;1) =e:
7.4We haveux+uy+2u=e 2ycos (x y) 2e 2ysin (x y) e 2ycos (x y)+
2e 2ysin (x y) = 0 andu(x;0) = sinx:
7.5(a) The general solution to this equation is u(x) =CwhereCis an
arbitrary constant.
(b) The general solution is u(x;y) =f(y) wherefis an arbitrary function of
y:
7.6(a) The general solution to this equation is u(x) =C1x+C2where
C1andC2are arbitrary constants.
(b) We have uy=f(y) wherefis an arbitrary function of y:Hence,u(x;y) =Ry
af(t)dt:
232 ANSWERS AND SOLUTIONS
7.7Letv(x;y) =y+ 2x:Then
ux=2fv(v) +g(v) + 2xgv(v)
uxx=4fvv(v) + 4gv(v) + 4xgvv(v)
uy=fv(v) +xgv(v)
uyy=fvv(v) +xgvv(v)
uxy=2fvv(v) +gv(v) + 2xgvv(v)
Hence,
uxx 4uxy+ 4uyy=4fvv(v) + 4gv(v) + 4xgvv(v)
8fvv(v) 4gv(v) 8xgvv(v)
+4fvv(v) + 4xgvv(v) = 0:
7.8utt=c2uxx:
7.9Letv=x+p(u)t:Using the chain rule we nd
ut=fvvt=fv(p(u) +puutt):
Thus
(1 tfvpu)ut=fvp:
If 1 tfvpu0 on anyt intervalIthenfvp0 onIwhich implies that
fv0 orp0 onI:But either condition will imply that tfvpu0 and
this will imply that 1 = 1 tfvpu= 0;a contradiction. Hence, we must have
1 tfvpu6= 0:In this case,
ut=fvp
1 tfvpu:
Likewise,
ux=fv(1 +puuxt)
or
ux=fv
1 tfvpu:
It follows that ut=p(u)ux:
Ifut= (sinu)uxthenp(u) = sinuso that the general solution is given by
u(x;t) =f(x+tsinu)
233
wherefis an arbitrary di erentiable function in one variable.
7.10u(x;y) =xf(x y) +g(x y):
7.11 Using integration by parts, we compute
ZL
0uxx(x;t)u(x;t)dx=ux(x;t)u(x;t)jL
x=0 ZL
0u2
x(x;t)dx
=ux(L;t)u(L;t) ux(0;t)u(0;t) ZL
0u2
x(x;t)dx
= ZL
0u2
x(x;t)dx0
Note that we have used the boundary conditions u(0;t) =u(L;t) = 0 and
the fact that u2
x(x;t)0 for allx2[0;L]:
7.12 (a) This can be done by plugging in the equations.
(b) Plug in.
(c) We have supfjun(x;0) 1j:x2Rg=1
nsupfjsinnxj:x2Rg=1
n:
(d) We have supfjun(x;t) 1j:x2Rg=en2t
n:
(e) We have lim t!1supfjun(x;t) 1j:x2R;t > 0g= limt!1en2t
n=1:
Hence, the solution is unstable and thus the problem is ill-posed.
7.13 (a)u(x;y) =x3+xy2+f(y);wherefis an arbitrary function.
(b)u(x;y) =x3y2
6+F(x) +g(y);whereF(x) =R
f(x)dx:
(c)u(x;t) =1
18e2x+3t+tRx
af(s)ds+Rx
ag(s)ds:
7.14 (b)u(x;y) =xf(y 2x) +g(y 2x):
7.15 We have
ut=cuv cuw
utt=c2uvv 2c2uwv+c2uww
ux=uv+uw
uxx=uvv+ 2uvw+uww
Substituting we nd uvw= 0 and solving this equation we nd uv=f(v)
andu(v;w) =F(v) +G(w) whereF(v) =R
f(v)dv:
234 ANSWERS AND SOLUTIONS
Finally, using the fact that v=x+ctandw=x ct; we get d'Alembert's
solution to the one-dimensional wave equation:
u(x;t) =F(x+ct) +G(x ct)
whereFandGare arbitrary di erentiable functions.
Section 8
8.1(a) Linear (b) Quasi-linear, nonlinear (c) Nonlinear (d) Semi-linear, non-
linear.
8.2Letw= 2x y:Thenux+ 2uy u=exf(w) + 2exfw(w) 2exfw(w)
exfw(w) = 0:
8.3We havexux yuy=x(pxy+xy
2pxy) yx2
2pxy=xpxy=u:Also,
u(y;y) =y2:
8.4We have yux+xuy= 2xysin (x2+y2) + 2xysin (x2+y2) = 0:More-
over,u(0;y) = cosy2:
8.5We have1
xux+1
yuy=1
x( x)+1
y(1+y) =1
y:Moreover,u(x;1) =1
2(3 x2):
8.63a 7b= 0:
8.7Plugu=av+winto the equation. Using the linearity of Land the
assumptions on vandw;obtain
L(u) =L(av+w) =aL(v) +L(w) = 0 +f=f
for any constant a:Therefore,usolves the nonhomogeneous equation for any
a:
8.8us+cu
a2+b2= 0:
8.9u(x;t) =1
2(x+y) +f(x y):
235
8.10 We have
ux= 4e 4xf(2x 3y) + 2e 4xf0(2x 3y)
uy= 3e 4xf0(2x 3y)
Thus,
3ux+ 2uy+ 12u= 12e 4xf(2x 3y) + 6e 4xf0(2x 3y)
6e 4xf0(2x 3y) + 12e 4xf(2x 3y) = 0:
8.11u(x;y) =f(ax bt)et
a:
8.12u(x;y) =f(bx ay):
8.13uw+u=f(v+cw;w ):
8.14vwv(v) =Aw(v):
Section 9
9.1u(x;t) = sin (x 3t):
9.2u(x;y) =e c(ax+by)
a2+b2f(bx ay):
9.3u(x;y) =xcos (2x y) +f(y 2x):
9.4The change of coordinates v=x+tandw=x treduces the
original equation to the equation uv=v+w
4whose solution is given by
u(v;w) =v2
8+wv
4+g(w) oru(x;t) =(x+t)2
8+x2 t2
4+g(x t):Butu(x;x) = 1
so that 1 =x2
2+g(0) org(0) = 1 x2
2which is impossible since g(0) is a
constant. Hence, the given initial value problem has no solution.
9.5u(x;t) =e 3t
1+(x 2t)2:
9.6u(x;t) =e3t
(x t)2+1
9
1
3t 1
9:
9.7Using the chain rule we nd wt=utet+uetandwx=uxet:Substi-
tuting these equations into the original equation we nd
wte t u+cwxe t+u= 0
236 ANSWERS AND SOLUTIONS
or
wt+cwx= 0
9.8u(x;y) =h(x y)
1 yh(x y):
9.9(a)w(x;t) is a solution to the equation follows from the principle of
superposition. Moreover, w(x;0) =u(x;0) v(x;0) =f(x) g(x):
(b)w(x;t) =f(x ct) g(x ct):
(c) From (b) we see that
max
x;tfju(x;t) v(x;t)jg= max
xfjf(x) g(x)jg:
Thus, small changes in the initial data produces small changes in the solu-
tion. Hence, the problem is a well-posed problem.
9.10
u(x;t) =
g
t x
c
e
cxifx<ct
0 if xct:
9.11u(x;t) = sin 2x 3t
2
:
9.12u(x;y) =1
2(x+y) +f(x y):
9.13 (a)a= 1; b= 0;c=B;andd= A(b)u(x;y) =f(Bx Ay)e C
Ax:
9.14u(x;y) =f(x y)e x:
9.15u(x;y) =f(x y)e x+x+y 2:
Section 10
10.1 The characteristics are hyperbolas: xy=k:
10.2 The characteristics are circles centered at the origin: x2+y2=k:
10.3 The characteristics are parallel lines with common slope equals to
1 :x y=k:
10.4f y
x;xe arctanu
= 0 wherefis an arbitrary di erentiable function.
237
10.5f(y+u;yln (y+u) x) = 0 oru= y+g(yln (y+u) x) where
fandgare arbitrary di erentiable functions.
10.6f y
x;u
x
= 0 oru=xg y
x
wherefandgare arbitrary di erentiable
functions.
10.7f y
x;u
xn
= 0 oru=xng y
x
wherefandgare arbitrary di eren-
tiable functions.
10.8f(x+y+z;xyu ) = 0 oru=g(x+y+z)
xywherefandgare arbitrary
di erentiable functions.
10.9f(xy;x4 u4 2xyu2) = 0;wherefis an arbitrary di erentiable func-
tion.
10.10f(x2+y2 u2;2xy+u2) = 0 where fis an arbitrary di erentiable
function.
10.11f(y
u;x2+y2+u2) = 0 where fis an arbitrary di erentiable func-
tion.
10.12u(x;y) =exf(y 2x):
10.13u(x;y) =f(y arctanx) for any di erentiable function f:The char-
acteristics are shown below.
238 ANSWERS AND SOLUTIONS
10.14u=exf(ye x) wherefis an arbitrary di erential function.
10.15 The characteristics are solutions to the DEdy
dx=x:Solving this ODE
we ndy=x2
2+C:
10.16u=e x2
2f(ye x) wherefis an arbitrary di erentiable function of
one variable.
10.17 Solvingdy
dx=x
yby the separation of variables we nd x2 y2=k;
wherekis a constant.
10.18 Solvingdy
dx=x
yby the separation of variables we nd x2 y2=k;
wherekis a constant.
Section 11
11.1u(x;y) =1 xy
x+y; x+y6= 0:
11.2u(x;y) = (x+y)(x2 y2):
11.3 2xyu+x2+y2 2u+ 2 = 0:
11.4u(x;y) = ln
x+ 1 y
x
:
239
11.5u(x;y) =f(xe y):
11.6u(t;x) =f(x at):
11.7u(x;y) =1
sec (x ay) y:
11.8u(x;y) =h
y (x 1)2
2 (x 1)
ex 1:
11.9u(x;y) =f(x uy):
11.10u(x;y) =y sin 1x:
11.11 (i)y=Cx2:The characteristics are parobolas in the plane centered
at the origin. See gure below.
(ii)u(x;y) =eyx 2:
(iii) In the rst case, we cannot substitute x= 0 intoyx 2(the argument
of the function f;above) because x 2is not de ned at 0. Similarly, in the
second case, we'd need to nd a function fso thatf(0) =h(x):Ifhis not
constant, it is not possible to satisfy this condition for all x2R:
(iv) All characteristics intersect at (0 ;0):Since the solution is constant along
any characteristic, if the solution is not exactly constant for all ( x;y);then
the limit of u(x;y) as (x;y)!(0;0) is di erent if we approach (0 ;0) along
di erent characteristics. Therefore, the method doesn't work at that point.
240 ANSWERS AND SOLUTIONS
11.12u(x;y) =eycos (x y):
11.13 (a)u=exf(ye x) wherefis an arbitrary di erential function.
(b) We want 2 = u(x;3x) =exf(3exe x) =exf(3):This equation is impossi-
ble so this Cauchy problem has no solutions.
(c) We want ex=exf(exe x) =)f(1) = 1:In this case, there are in nitely
many solutions to this Cauchy problem, namely, u(x;y) =exf(ye x) where
fis an arbitrary function satisfying f(1) = 1:
11.14u(x;y) = 1 + 2ex2
2e (4x y)2
2:
11.15 The Cauchy problem has no solutions.
11.16 (a) The characteristics satisfy the ODEdy
dx=x
y:Solving this equa-
tion we nd x2 y2=C:Thus, the characteristics are hyperbolas.
(b)
(c) The general solution to the PDE is u(x;y) =f(x2 y2) wherefis
an arbitrary di erentiable function. Since u(0;y) =e y2we ndf(y) =ey:
Hence,u(x;y) =ex2 y2:
(d) This solution is only de ned in the region covered by characteristics that
cross theyaxis:y2 x2>0:The solution in the region y2 x2<0 can be
any function of the form u(x;y) =f(x2 y2):
241
11.17 (a) Solving the ODEdy
dx=ywe nd the characteristics ye x=C:
Thus,u(x;y) =f(ye x):Ifu(x;0) = 1 then we choose fto be any arbitrary
di erentiable function satisfying f(0) = 1:
(b) The line y= 0 is a characteristic so that uhas to be constant there.
Hence, there is no solution satisfying the condition u(x;0) =x:
Section 12
12.1 (a) Hyperbolic (b) Parabolic (c) Elliptic.
12.2 (a) Ellitpic (b) Parabolic (c) Hyperbolic.
12.3The PDE is of hyperbolic type if 4 y2(x2+x+ 1)>0:This is true for
ally6= 0:Graphically, this is the xy plane with the x axis removed,
The PDE is of parabolic type if 4 y2(x2+x+ 1) = 0:Sincex2+x+ 1>0
for allx2R;we must have y= 0:Graphically, this is x axis.
The PDE is of elliptic type if 4 y2(x2+x+ 1)<0 which can not happen.
12.4 We have
ux(x;t) = sinxsint;
uxx(x;t) = cosxsint;
ut(x;t) = cosxcost;
utt(x;t) = cosxsint:
Thus,
uxx(x;t) = cosxsint=utt(x;t);
u(x;0) = cos xsin 0 = 0;
ut(x;0) = cos xcos 0 = cosx;
ux(0;t) = sin 0 sint= 0:
12.5 (a) Quasi-linear (b) Semi-linear (c) Linear (d) Nonlinear.
242 ANSWERS AND SOLUTIONS
12.6 We have
ux=2x
x2+y2
uxx=2y2 2x2
(x2+y2)2
uy=2y
x2+y2
uyy=2x2 2y2
(x2+y2)2
Plugging these expressions into the equation we nd uxx+uyy= 0:Similar
argument holds for the second part of the problem.
12.7 Multiplying the equation by uand integrating, we obtain
ZL
0u2(x)dx=ZL
0uuxx(x)dx
=[u(L)ux(L) u(0)ux(0)] ZL
0u2
x(x)dx
=
kLu(L)2+k0u(0)2+ZL
0u2
x(x)dx
For > 0;becausek0;kL>0;the right-hand side is nonpositive and the
left-hand side is nonnegative. Therefore, both sides must be zero, and there
can be no solution other than u0;which is the trivial solution.
12.8 Substituteu(x;y) =f(x)g(y) into the left side of the equation to obtain
f(x)g(y)(f(x)g(y))xy=f(x)g(y)f0(x)g0(y):Now, substitute the same thing
into the right side to obtain ( f(x)g(y))x(f(x)g(y))y=f0(x)g(y)f(x)g0(y) =
f(x)g(y)f0(x)g0(y):So the sides are equal, which means f(x)g(y) is a solution.
12.9 We have
(un)xx= n2sinnxsinhnyand (un)yy=n2sinnxsinhny
Hence, un= 0:
12.10u(x;y) =x2y2
4+F(x) +G(y);whereF(x) =R
f(x)dx:
243
12.11 (a) We have A= 2; B= 4; C= 7 soB2 4AC= 16 56 = 40<0:
So this equation is elliptic everywhere in R2:
(b) We have A= 1; B= 2 cosx; C = sin2xsoB2 4AC= 4 cos2x+
4 sin2x= 4>0:So this equation is hyperbolic everywhere in R2:
(c) We have A=y; B = 2(x 1); C = (y+ 2) soB2 4AC =
4(x 1)2+ 4y(y+ 2) = 4[(x 1)2+ (y+ 1)2 4]:The equation is parabolic if
(x 1)2+ (y+ 1)2= 4:It is hyperbolic if ( x 1)2+ (y+ 1)2>4 and elliptic
if (x 1)2+ (y+ 1)2<4:
12.12 Using the chain rule we nd
ut(x;t) =1
2(cf0(x+ct) cf0(x ct)) +1
2c[g(x+ct)(c) g(x ct)( c))
=c
2(f0(x+ct) f0(x ct)) +1
2(g(x+ct) +g(x ct))
utt=c2
2(f00(x+ct) +f00(x ct)) +c
2(g0(x+ct) g0(c xt))
ux(x;t) =1
2(f0(x+ct) +f0(x ct)) +1
2c[g(x+ct) g(x ct)]
uxx(x;t) =1
2(f00(x+ct) +f00(x ct)) +1
2c[g0(x+ct) g0(x ct)]
By substitutition we see that c2uxx=utt:Moreover,
u(x;0) =1
2(f(x) +f(x)) +1
2cZx
xg(s)ds=f(x)
and
ut(x;0) =g(x):
12.13 (a) 1 + 4x2y>0;(b) 1 + 4x2y= 0;(c) 1 + 4x2y<0:
12.14u(x;y) =f(y 3x) +g(x+y):
12.15u(x;y) =f(y 3x) +g(x+y) =10x2+y2 7xy+6
6:
Section 13
244 ANSWERS AND SOLUTIONS
13.1 Letz(x;t) = v(x;t) + w(x;t):Then we have
c2zxx=c2 vxx+c2 wxx
= vtt+ vtt
=ztt:
13.2 Indeed we have c2uxx(x;t) = 0 =utt(x;t):
13.3u(x;t) = 0:
13.4u(x;t) =1
2(cos (x 3t) + cos (x+ 3t)):
13.5u(x;t) =1
2h
1
1+(x+t)2+1
1+(x t)2i
:
13.6u(x;t) = 1 +1
8[sin (2x+ 4t) sin (2x 4t)]:
13.7
u(x;t) =8
>><
>>:1 ifx 5t<0 andx+ 5t<0
1
2ifx 5t<0 andx+ 5t>0
1
2ifx 5t>0 andx+ 5t<0
0 ifx 5t>0 andx+ 5t>0
13.8u(x;t) =1
2[e (x+ct)2+e (x ct)2] +t
2+1
4ccos (2x) sin (2ct):
13.9 Just plug the translated/di erentiated/dialated solution into the wave
equation and check that it is a solution.
13.10v(r) =Acos (nr) +Bsin (nr):
13.11u(x;t) =1
2[ex ct+ex+ct+1
c(cos (x ct) cos (x+ct))]:
245
13.12 (a) We have
dE
dt(t) =ZL
0ututtdx+ZL
0c2uxuxtdx
=ZL
0ututtdx+c2ut(L;t)ux(L;t) c2ut(0;t)ux(0;t) c2ZL
0utuxxdx
=c2ut(L;t)ux(L;t) c2ut(0;t)ux(0;t) +ZL
0ut(utt c2uxx)dx
=c2(ut(L;t)ux(L;t) ut(0;t)ux(0;t))
sinceutt c2uxx= 0:
(b) Since the ends are xed, we have ut(0;t) =ut(L;t) = 0:From (a) we
have
dE
dt(t) =c2(ut(L;t)ux(L;t) ut(0;t)ux(0;t)) = 0:
(c) Assuming free ends boundary conditions, that is ux(0;t) =ux(L;t) = 0;
we nddE
dt(t) = 0:
13.13 Using the previous exercise, we nd
dE
dt(t) = dZL
0(ut)2dx:
The right-hand side is nonpositive, so the energy either decreases or is con-
stant. The latter case can occur only if ut(x;t) is identically zero, which
means that the string is at rest.
13.14 (a) By the chain rule we have ut(x;t) = cR0(x ct) andutt(x;t) =
c2R00(x ct):Likewise,ux(x;t) =R0(x ct) anduxx=R00(x ct):Thus,
utt=c2uxx:
(b) We have
1
2ZL
0(ut)2dx=ZL
0c2
2[R0(x ct)]2dx=ZL
0c2
2(ux)2dx:
13.15u(x;t) =x2+ 4t2+1
4sin 2xsin 4t:
Section 14
246 ANSWERS AND SOLUTIONS
14.1 Letz(x;t) = u(x;t) + v(x;t):Then we have
kzxx=k uxx+k vxx
= ut+ vt
=zt:
14.2 Indeed we have kuxx(x;t) = 0 =ut(x;t):
14.3u(x;t) =T0+TL T0
Lx:
14.4 Letube the solution to (14.1) that satis es u(0;t) =u(L;t) = 0:Let
w(x;t) be the time independent solution to (14.1) that satis es w(0;t) =T0
andw(L;t) =TL:That is,w(x;t) =T0+TL T0
Lx:From Exercise 14.1,
the function u(x;t) =u(x;t) +w(x;t) is a solution to (14.1) that satis-
esu(0;t) =T0andu(L;t) =TL:
14.5u(x;t) = 0:
14.6 Substituting u(x;t) =X(x)T(t) into (14.1) we obtain
kX00
X=T0
T:
SinceXonly depends on xandTonly depends on t;we must have that
there is a constant such that
kX00
X=andT0
T=:
This gives the two ordinary di erential equations
X00
kX= 0 andT0 T= 0:
14.7 (a) Letting =
k>0 we obtain the ODE X00 X= 0 whose general
solution is given by X(x) =Aexp +Be xp for some constants AandB:
(b) The condition u(0;t) = 0 implies that X(0) = 0 which in turn implies
A+B= 0:Likewise, the condition u(L;t) = 0 implies Aep
L +Be Lp = 0:
Hence,A(eLp e Lp ) = 0:
(c) IfA= 0 thenB= 0 andu(x;t) is the trivial solution which contradicts
247
the assumption that uis non-trivial. Hence, we must have A6= 0:
(d) Using (b) and (c) we obtain eLp =e Lp ore2Lp = 1:This equa-
tion is impossible since 2 Lp > 0:Hence, we must have < 0 so that
X(x) =Acos (xp ) +Bsin (xp :
14.8 (a)Now, write =q
k:Then we obtain the equation X00+ 2X= 0
whose general solution is given by
X(x) =c1cos x+c2sin x:
(b) UsingX(0) = 0 we obtain c1= 0:Sincec26= 0 we must have sin L= 0:
Thus,= kn22
L2;wherenis an integer.
14.9 For each integer n0 we have un(x;t) =cn
T(0)T(0)ekn22
L2tsin n
L
x
is a solution to (14.1). By superposition, u(x;t) is also a solution to (14.1).
Moreover,u(0;t) =u(L;t) = 0 since un(0;t) =un(L;t) = 0:
14.10 (i)u(0;t) = 0 andu(a;t) = 100 for t>0:
(ii)ux(0;t) =ux(a;t) = 0 fort>0:
14.11 Solving this problem we nd u(x;t) =e tsinx:We have
E(t) =Z
0[e 2tsin2x+e 2tcos2x]dx=Z
0e 2tdx=e 2t:
Thus,E0(t) = 2e 2t<0 for allt>0:
14.12E(t) =RL
0f(x)dx+ (1 + 4L)t:
14.13v(x) =x+ 2:
14.14 (a)v(x) =T
Lx:
(b)v(x) =T:
(c)v(x) = x+T:
14.15 (a)E(t) =RL
0cu(x;t)dx:
(b) We integrate the equation in xfrom 0 toL:
ZL
0cut(x;t)dx=ZL
0Kuxxdx=Kux(x;t)jL
0= 0;
248 ANSWERS AND SOLUTIONS
sinceux(0;t) =ux(L;t) = 0:The left-hand side can also be written as
d
dtZL
0cu(x;t)dx=E0(t):
Thus, we have shown that E0(t) = 0 so that E(t) is constant.
14.16 (a) The total thermal energy is
E(t) =ZL
0u(x;t)dx:
We have
dE
dt=ZL
0ut(x;t)dx=uxjL
0+ZL
0xdx= (7 ) +L2
2:
(b) The steady solution (equilibrium) is possible if the right-hand side van-
ishes:
(7 ) +L2
2= 0
Solving this equation for we nd = 7 +L2
2:
(c) By integrating the equation uxx+x= 0 we nd the steady solution
u(x) = x3
6+C1x+C2
From the condition ux(0) = we ndC1= :The steady solution should
also have the same value of the total energy as the initial condition. This
means ZL
0
x3
6+ x+C2
dx=ZL
0f(x)dx=E(0):
Performing the integration and then solving for C2we nd
C2=1
LZL
0f(x)dx+L3
24 L
2:
Therefore, the steady-state solution is
u(x) =1
LZL
0f(x)dx+L3
24 L
2+ x x3
6:
249
Section 15
15.1 (a) We have ( fg)(x+T) =f(x+T)g(x+T) =f(x)g(x) = (fg)(x):
(b) We have ( c1f+c2g)(x+T) =c1f(x+T)+c2g(x+T) =c1f(x)+c2g(x) =
(c1f+c2g)(x):
15.2 (a) Forn6=mwe have
ZL
Lsinm
Lx
sinn
Lx
dx= 1
2ZL
L
cos(m+n)
Lx
cos(m n)
Lx
dx
= 1
2L
(m+n)sin(m+n)
Lx
L
(m n)sin(m n)
LxL
L
=0
where we used the trigonometric identiy
sinasinb=1
2[ cos (a+b) + cos (a b)]:
(b) Forn6=mwe have
ZL
Lcosm
Lx
sinn
Lx
dx=1
2ZL
L
sin(m+n)
Lx
sin(m n)
Lx
dx
=1
2
L
(m+n)cos(m+n)
Lx
+L
(m n)cos(m n)
LxL
L
=0
where we used the trigonometric identiy
cosasinb=1
2[sin (a+b) sin (a b)]:
15.3 (a) L (b) L (c) 0.
250 ANSWERS AND SOLUTIONS
15.4
a0=1
Z
f(x)dx= 0
an=1
Z
f(x) cosnxdx
= Z0
cosnxdx +Z
0cosnxdx = 0
bn=1
Z
f(x) sinnxdx
= Z0
sinnxdx +Z
0sinnxdx
=2
n[1 ( 1)n]
15.5f(x) = 1
6+P1
n=14
(n)2( 1)ncos (nx):
15.6f(x) =P1
n=12
n
cos n
2
( 1)n
sin nx
2
:
15.7f(x) =P1
n=14
(n)2[1 ( 1)n] cos n
2x
:
15.8 Since the sided limits at the point of discontinuity x= 0 do not exist,
the function is not piecewise continuous in [ 1;1]:
15.9 De ne the function
g(a) =ZL+a
L+af(x)dx:
Using the fundamental theorem of calculus, we have
dg
da=d
daZL+a
L+af(x)dx
=f(L+a) f( L+a) =f( L+a+ 2L) f( L+a)
=f( L+a) f( L+a) = 0
Hence,gis a constant function, and in particular we can write g(a) =g(0)
for alla2Rwhich gives the desired result.
251
15.10 (i)f(x) =10
3+P1
n=1
1
nsin 2n
3
cos 2nx
3
1
n
cos 2n
3
+ 1
sin 2nx
3
:
(ii) Using the theorem discussed in class, because this function and its deriva-
tive are piecewise continuous, the Fourier series will converge to the function
at each point of continuity. At any point of discontinuity, the Fourier series
will converge to the average of the left and right limits.
(iii)
.
15.11 (a)a0= 2; an=bn= 0 forn2N:
(b)a0= 4; an= 0; b1= 1;andbn= 0:
(c)a0= 1; an= 0; bn=1
n[1 ( 1)n]; n2N:
(d)a0=an= 0; bn=2L
n( 1)n+1; n2N:
15.12 1
15.13an= 0 for alln2N:
15.14f(0 )+f(0+)
2= +
2= 0:
15.15 (a)f(x) =3
2+2
P1
n=1sin (2n 1)x
2n 1:
(b)P1
n=1( 1)n+1
2n 1=
4:
Section 16
16.1f(x) = 0:
16.2
252 ANSWERS AND SOLUTIONS
16.3
16.4
253
16.5f(x) =
4+P1
n=12
n2[2 cos (n=2) 1 ( 1)n] cosnx:
16.6f(x) =
2+P1
n=12
n2[( 1)n 1] cosnx:
16.7f(x) =P1
n=12
n[1 ( 1)n] sinnx:
16.8f(x) =2
P1
n=1n
1 ( 1)n
n2 1
sinnx:
16.9f(x) =1
2(e2 1) +P1
n=14[( 1)ne2 1]
4+n22cos (nx):
16.10 (a) Iff(x) = sin 2
Lx
thenbn= 0 ifn6= 2 andb2= 1:
(b) Iff(x) = 1 then
bn=2
LZL
0sinn
Lx
dx=2
n[1 ( 1)n]:
(c) Iff(x) = cos
Lx
then
b1=2
LZL
0cos
Lx
sin
Lx
dx= 0
254 ANSWERS AND SOLUTIONS
and forn6= 1 we have
bn=2
LZL
0cos
Lx
sinn
Lx
dx
=1
22
LZL
0h
sinx
L
(1 +n) sinx
L
(1 n)i
dx
=1
L
L
(1 +n)cosx
L
(1 +n) +L
(1 n)cosx
L
(1 n)L
0
=2n
(n2 1)[1 ( 1)n]:
16.11 (a)a0= 10 anda1= 1;andan= 0 forn6= 1:
(b)a0=Landan=2L
(n)2[( 1)n 1]; n2N:
(c)a0= 1 andan=2
nsin n
2
; n2N:
16.12 By de nition of Fourier sine coecients,
bn=2
LZL
0f(x) sinn
Lx
dx
The symmetry around x=L
2can be written as
fL
2+x
=fL
2 x
for allx2R:To use this symmetry it is convenient to make the change of
variablex L
2=uin the above integral to obtain
bn=ZL
2
L
2fL
2+u
sinn
LL
2+u
du:
Sincef L
2+u
is even inuand forneven sinn
L L
2+u
= sin nu
L
is
odd inu;the integrand of the above integral is odd in uforneven. Since
the intergral is from L
2toL
2we must have b2n= 0 forn= 0;1;2;
16.13 By de nition of Fourier cosine coecients,
an=2
LZL
0f(x) cosn
Lx
dx
255
The anti-symmetry around x=L
2can be written as
fL
2 y
= fL
2+y
for ally2R:To use this symmetry it is convenient to make the change of
variablex=L
2+yin the above integral to obtain
an=ZL
2
L
2fL
2+y
cosn
LL
2+y
dy:
Sincef L
2+y
is odd inyand forneven cosn
L L
2+y
=cos ny
L
is
even iny;the integrand of the above integral is odd in yforneven. Since
the intergral is from L
2toL
2we must have a2n= 0 for alln= 0;1;2;:
16.14 sin nx
L
=2
2
P1
n=21 ( 1)n
n2 1cos nx
L
:
16.15 (a)
(b)a0=2
2R2
0f(x)dx= 3:
256 ANSWERS AND SOLUTIONS
(c) We have
an=2
2Z2
0f(x) cosnx
2
dx
=Z1
0cosnx
2
dx+Z2
12 cosnx
2
dx
=2
nsinnx
2 1
0+ 22
nsinnx
2 2
1
= 2
nsinn
2
:
(d)bn= 0 sincef(x) sin nx
2
is odd in 2x2:
(e)
f(x) =3
2+1X
n=1
2
nsinn
2
cosnx
2
:
Section 17
17.1 We look for a solution of the form u(x;y) =X(x)Y(y):Substituting in
the given equation, we obtain
X00Y+XY00+XY = 0:
AssumingX(x)Y(y) is nonzero, dividing for X(x)Y(y) and subtract both
sides forX00(x)
X(x);we nd:
X00(x)
X(x)=Y00(y)
Y(y)+:
The left hand side is a function of xwhile the right hand side is a function
ofy:This says that they must equal to a constant. That is,
X00(x)
X(x)=Y00(y)
Y(y)+=:
whereis a constant. This results in the following two ODEs
X00+X= 0 andY00+ ( )Y= 0:
257
If>0 and >0 then
X(x) =Acosx+Bsinx
Y(y) =Ccos ( )y+Dsin ( )y
If>0 and <0 then
X(x) =Acosx+Bsinx
Y(y) =Ce p
( )y+Dep
( )y
If=>0 then
X(x) =Acosx+Bsinx
Y(y) =Cy+D
If=<0 then
X(x) =Ae p
x+Bep
x
Y(y) =Cy+D
If<0 and >0 then
X(x) =Ae p
x+Bep
x
Y(y) =Ccos ( )y+Dsin ( )y
If<0 and <0 then
X(x) =Ae p
x+Bep
x
Y(y) =Ce p
( )y+Dep
( )y:
17.2 Let's assume that the solution can be written in the form u(x;t) =
X(x)T(t):Substituting into the heat equation we obtain
X00
X=T0
kT:
SinceXonly depends on xandTonly depends on t;we must have that
there is a constant such that
X00
X=andT0
kT=:
258 ANSWERS AND SOLUTIONS
This gives the two ordinary di erential equations
X00 X= 0 andT0 kT = 0:
Next, we consider the three cases of the sign of :
Case 1:= 0
In this case, X00= 0 andT0= 0:Solving these equations we nd X(x) =
ax+bandT(t) =c:
Case 2:>0
In this case, X(x) =Aep
x+Be p
xandT(t) =Cekt:
Case 3:<0
In this case, X(x) =Acosp
x+Bsinp
xand andT(t) =Cekt:
17.3r2R00(r) +rR0(r) R(r) = 0 and 00() +() = 0:
17.4X00= (2 +)X; T00=T;X (0) = 0; X(1) = 0:
17.5X00 X= 0; T0=kT;X0(0) = 0 =X0(L):
17.6u(x;t) =Ce(x t):
17.7 5X000 7X00 X= 0 and 3Y00 Y0= 0:
17.8u(x;y) =Cex y
:
17.9u(x;y) =Cexy:
17.10 We look for a solution of the form u(x;y) =X(x)T(t):Substitut-
ing in the wave equation, we obtain
X00(x)T(t) X(x)T00(t) = 0:
AssumingX(x)T(t) is nonzero, dividing for X(x)T(t) we nd:
X00(x)
X(x)=T00(t)
T(t):
259
The left hand side is a function of xwhile the right hand side is a function
oft:This says that they must equal to a constant. That is,
X00(x)
X(x)=T00(t)
T(t)=
whereis a constant. This results in the following two ODEs
X00 X= 0 andT00 T= 0:
The solutions of these equations depend on the sign of :
If>0 then the solutions are given
X(x) =Aep
x+Be p
x
T(t) =Cep
t+De p
t
whereA;B;C; andDare constants. In this case,
u(x;t) =k1ep
(x+t)+k2ep
(x t)+k3e p
(x+t)+k4e p
(x t):
If= 0 then
X(x) =Ax+B
T(t) =Ct+D
whereA;B; andCare arbitrary constants. In this case,
u(x;t) =k1xt+k2x+k3t+k4:
If<0 then
X(x) =Acosp
x+Bsinp
x
T(t) =Acosp
t+Bsinp
t
whereA;B;C; andDare arbitrary constants. In this case,
u(x;t) =k1cosp
xcosp
t+k2cosp
xsinp
t
+k3sinp
xcosp
t+k4sinp
xsinp
t:
17.11 (a)u(r;t) =R(r)T(t); T0(t) =kT; r (rR0)0=R:
260 ANSWERS AND SOLUTIONS
(b)u(x;t) =X(x)T(t); T0=T; kX00 ( +)X= 0:
(c)u(x;t) =X(x)T(t); T0=T; kX00 aX0=X:
(d)u(x;t) =X(x)Y(y); X00=X; Y00= Y:
(e)u(x;t) =X(x)T(t); T0=kT; X0000=X:
17.12u(x;y) =Ce(x+y):
17.13X00=X; Y0 Y00+Y=Y:
Section 18
18.1u(x;t) = sin
2x
e 2k
4t+ 3 sin 5
2x
e 252k
4t:
18.2u(x;t) =8d
3P1
n=11
(2n 1)3sin
(2n 1)
Lx
e k(2n 1)22
L2t:
18.3u(x;t) =2
4
P1
n=11
(4n2 1)cos 2n
Lx
e k4n22
L2t:
18.4u(x;t) =P1
n=1Cnsin n
Lx
e n22
L2twhere
Cn=8
<
: 4
nn= 2;6;10;
0n= 4;8;12;
6
nnis odd.
18.5u(x;t) = 6 sin 9
Lx
e 812
L2t:
18.6u(x;t) =1
2+P1
n=1Cncos n
Lx
e n22
L2twhere
Cn=8
<
: 2
nn= 1;5;9;
2
nn= 3;7;11;
0nis even
18.7u(x;t) = 6 + 4 cos 3
Lx
e 92
L2t:
18.8u(x;t) = 3 cos 8
Lx
e 642
L2t:
18.9
u(x;t) =1X
n=0ancosn
Lx
e
1+n22
L2
t:
261
Ast!1; e
1+n22
L2
t!0 for eachn2N:Hence,u(x;t)!0:
18.10 (b) We have
E0(t) =2Z1
0w(x;t)wt(x;t)dx
=2Z1
0w(x;t)[wxx(x;t) w(x;t)]dx
= 2w(x;t)wx(x;t)j1
0 2Z1
0w2
x(x;t)dx+Z1
0w2(x;t)dx
= 2Z1
0w2
x(x;t)dx+Z1
0w2(x;t)dx
0
Hence,Eis decreasing, and 0 E(t)E(0) for allt>0:
(c) Sincew(x;0) = 0;we must have E(0) = 0:Hence,E(t) = 0 for all t0:
This implies that w(x;t) = 0 for all t >0 and all 0 < x < 1:Therefore
u1(x;t) =u2(x;t):This means that the given problem has a unique solution.
18.11 (a)u(0;t) = 0 andux(1;t) = 0:
(b) Let's assume that the solution can be written in the form u(x;t) =
X(x)T(t):Substituting into the heat equation we obtain
X00
X=T0
kT:
SinceXonly depends on xandTonly depends on t;we must have that
there is a constant such that
X00
X=andT0
kT=:
This gives the two ordinary di erential equations
X00 X= 0 andT0 kT = 0:
As far as the boundary conditions, we have
u(0;t) = 0 =X(0)T(t) =)X(0) = 0
and
ux(1;t) = 0 =X0(1)T(t) =)X0(1) = 0:
262 ANSWERS AND SOLUTIONS
Note thatTis not the zero function for otherwise u0 and this contradicts
our assumption that uis the non-trivial solution.
(c) We have X0=p
cosp
xandX00=sinp
x:Thus,X00 X= 0:
MoreoverX(0) = 0:Now,X0(1) = 0 implies cosp
= 0 orp
=
n 1
2
; n2N:Hence,=
n 1
222:
18.12 (a) Let's assume that the solution can be written in the form u(x;t) =
X(x)T(t):Substituting into the heat equation we obtain
X00
X=T0
kT:
Since the LHS only depends on xand the RHS only depends on t;there must
be a constant such that
X00
X=andT0
kT=:
This gives the two ordinary di erential equations
X00 X= 0 andT0 kT = 0:
As far as the boundary conditions, we have
u(0;t) = 0 =X(0)T(t) =)X(0) = 0
and
u(L;t) = 0 =X(L)T(t) =)X(L) = 0:
Note thatTis not the zero function for otherwise u0 and this contradicts
our assumption that uis the non-trivial solution.
Next, we consider the three cases of the sign of :
Case 1:= 0
In this case, X00= 0:Solving this equation we nd X(x) =ax+b:Since
X(0) = 0 we nd b= 0:SinceX(L) = 0 we nd a= 0:Hence,X0 and
u(x;t)0:That is,uis the trivial solution.
Case 2:>0
In this case, X(x) =Aep
x+Be p
x:Again, the conditions X(0) =X(L) =
0 implyA=B= 0 and hence the solution is the trivial solution.
263
Case 3:<0
In this case, X(x) =Acosp
x+Bsinp
x:The condition X(0) = 0
impliesA= 0:The condition X(L) = 0 implies Bsinp
L= 0:We must
haveB6= 0 otherwise X(x) = 0 and this leads to the trivial solution. Since
B6= 0, we obtain sinp
L= 0 orp
L=nwheren2N:Solving for
we nd= n22
L2:Thus, we obtain in nitely many solutions given by
Xn(x) =Ansinn
Lx; n2N:
Now, solving the equation
T0 kT = 0
by the method of separation of variables we obtain
Tn(t) =Bne n22
L2kt; n2N:
Hence, the functions
un(x;t) =Cnsinn
Lx
e n22
L2kt; n2N
satisfyut=kuxxand the boundary conditions u(0;t) =u(L;t) = 0:
Now, in order for these solutions to satisfy the initial value condition u(x;0) =
6 sin 9x
L
;we invoke the superposition principle of linear PDE to write
u(x;t) =1X
n=1Cnsinn
Lx
e n22
L2kt: (1)
To determine the unknown constants Cnwe use the initial condition u(x;0) =
6 sin 9x
L
in (1) to obtain
6 sin9x
L
=1X
n=1Cnsinn
Lx
:
By equating coecients we nd C9= 6 andCn= 0 ifn6= 9:Hence, the
solution to the problem is given by
u(x;t) = 6 sin9x
L
e 812
L2kt:
264 ANSWERS AND SOLUTIONS
(b) Similar to (a), we nd
u(x;t) = 3 sin
Lx
e 2kt
L2 sin3
Lx
e 92kt
L2
18.13u(x;t) = cos x
L
e pi2kt
L2+ 4 cos 5x
L
e 25pi2kt
L2:
(b)u(x;t) = 5:
18.14u(x;t) = 6 sinxe 8t:
Section 19
19.1u(x;y) =P1
n=1Bnsin n
by
sinh n
bx
where
Bn=2
bZb
0f2(y) sinn
by
dyh
sinhn
bai 1
:
19.2u(x;y) =P1
n=1Bnsinn
axsinh n
a(y b)
where
Bn=2
aZa
0g1(x) sinn
ax
dx
[sinh
n
ab
] 1:
19.3u(x;y) = 2xy+3
sinhsinxsinhy:
19.4 Ifu(x;y) =x2 y2thenuxx= 2 anduyy= 2 so that u= 0:
Ifu(x;y) = 2xythenuxx=uyy= 0 so that u= 0:
19.5
u(x;y) =1X
n=1[Ancoshn
Ly
+Bnsinhn
Ly
] sinn
Lx:
where
An=2
LZL
0(f1(x) +f2(x)) sinn
Lxdx
coshnH
2L 1
and
Bn=2
LZL
0(f2(x) f1(x)) sinn
Lxdx
sinhnH
2L 1
265
19.6 (a) Di erentiating term by term with respect to xwe nd
ux+ivx=1X
n=0nan(x+iy)n 1:
Likewise, di erentiating term by term with respect to ywe nd
uy+ivy=1X
n=0nani(x+iy)n 1:
Multiply this equation by iwe nd
iuy+vy=1X
n=0nan(x+iy)n 1:
Hence,ux+ivx=vy iuywhich implies ux=vyandvx= uy:
(b) We have uxx= (vy)x= (vx)y= uyyso that u= 0:Similar argument
for v= 0:
19.7 Polar and Cartesian coordinates are related by the expressions x=
rcosandy=rsinwherer= (x2+y2)1
2and tan=y
x:Using the chain
rule we obtain
ux=urrx+ux= cosur sin
ru
uxx=uxrrx+uxx
=
cosurr+sin
r2u sin
rur
cos
+
sinur+ cosur cos
ru sin
ru
sin
r
uy=urry+uy= sinur+cos
ru
uyy=uyrry+uyy
=
sinurr cos
r2u+cos
rur
sin
+
cosur+ sinur sin
ru+cos
rucos
r
266 ANSWERS AND SOLUTIONS
Substituting these equations into (21.1) we obtain the dersired equation.
19.8u(x;y) =u1(x;y) +u2(x;y) +u3(x;y) +u4(x;y) where
u1(x;y) = 0
u2(x;y) =1X
n=1"
2
n( 1)n
sinh 3n
2#
sinn
2xsinhn
2y
u3(x;y) =1
sinh 8
3sinh4(x 2)
3
sin4
3y
u4(x;y) =1X
n=114(1 ( 1)n)
nsinh 2n
3sinn
3y
sinhn
3x
:
19.9
u(x;y) =4
sinh L
2H
sinhx
2H
sinh(x L)
2H
cosy
2H:
19.10u(x;t) =A0+P1
n=1Ane pnxcospnywhere
A0=1
HZH
0f(y)dy
An=2
HZH
0f(y) cosn
Hydy:
19.11u(x;y) =20
Y1(H)Y1(y) sin x
L
5
Y3(H)sin 3x
L
:
19.12u(x;y) = sin (2x)e 2y:
19.13u(x;y) =y:
19.14u(x;y) =1
2x2 1
2y2 axby+CwhereCis an arbitrary constant.
19.15u(x;y) =2 cosh 3ysin 3x
cosh 6 5 cosh 10ysin 10x
cosh 20:
Section 20
267
20.1u(r;) = 3r5sin 5:
20.2u(r;) =
4+P1
n=1rnh
1 ( 1)n
n2cosn+sinn
ni
:
20.3u(r;) =C0+r2cos 2:
20.4 Substituting C0;An;andBninto the right-hand side of u(r;) we nd
u(r;) =1
2Z2
0f( )d +1X
n=1rn
anZ2
0f( ) [cosn cosn+ sinn sinn]d
=1
2Z2
0f( )"
1 + 21X
n=1r
an
cosn( )#
d :
20.5 (a) We have eit= cost+isintande it= cost isint:The result
follows by adding these two equalities and dividing by 2.
(b) This follows from the fact that
cosn( ) =1
2(ein( )+e in( )):
(c) We havejq1j=r
aq
cos ( )2+ sin ( )2=r
a<1 since 0<r<a: A
similar argument shows that jq2j<1:
20.6 (a) The rst sum is a convergent geometric series with ratio q1and
sum
1X
n=1r
an
ein( )=r
aei( )
1 q1
=rei( )
a rei( )
268 ANSWERS AND SOLUTIONS
Similar argument for the second sum.
(b) We have
1 + 21X
n=1r
an
cosn( ) =1 +rei( )
a rei( )
+re i( )
a re i( )
=1 +r
ae i( ) r+r
ae i( ) r
=1 +r
acos ( ) r aisin ( )
+r
acos ( ) r+aisin ( )
=1 +r[acos ( ) r+aisin ( )]
a2+ 2arcos ( ) +r2
+r[acos ( ) r aisin ( )]
a2 2arcos ( ) +r2
=a2 r2
a2 2arcos ( ) +r2:
20.7 We have
u(r;) =1
2Z2
0f( )"
1 + 21X
n=1r
an
cosn( )#
d
=1
2Z2
0f( )a2 r2
a2 2arcos ( ) +r2d
=a2 r2
2Z2
0f( )
a2 2arcos ( ) +r2d :
20.8u(r;) = 2P1
n=1( 1)n+1rnsinn
n:
20.9 (a) Di erentiating u(r;t) =R(r)T(t) with respect to randtwe nd
utt=RT00andur=R0Tandurr=R00T:
Substituting these into the given PDE we nd
RT00=c2
R00T+1
rR0T
269
Dividing both sides by c2RTwe nd
1
c2T00
T=R00
R+1
rR0
R:
Since the RHS of the above equation depends on ronly, and the LHS depends
ontonly, they must equal to a constant :
(b) The given boundary conditions imply
u(a;t) = 0 =R(a)T(t) =)R(a) = 0
u(r;0) =f(r) =R(r)T(0)
ut(r;0) =g(r) =R(r)T0(0):
If= 0 thenR00+1
rR0= 0 and this implies R(r) =Clnr:Using the condition
R(a) = 0 we nd C= 0 so that R(r) = 0 and hence u0:If > 0 then
T00 c2T= 0:This equation has the solution
T(t) =Acos (cp
t) +Bsin (cp
t):
The condition u(r;0) =f(r) implies that A=f(r) which is not possible.
Hence,<0:
20.10 (a) Follows from the gure and the de nitions of trigonometric func-
tions in a right triangle.
(b) The result follows from equation (20.1).
20.11 By the maximum principle we have
min
(x;y)2@
u(x;y)u(x;y)max
(x;y)2@
u(x;y);8(x;y)2
But min (x;y)2@
u(x;y) =u(1;0) = 1 and max (x;y)2@
u(x;y) =u( 1;0) = 3:
Hence,
1u(x;y)3
and this implies that u(x;y)>0 for all (x;y)2
:
20.12 (i)u(1;0) = 4 (ii)u( 1;0) = 2:
270 ANSWERS AND SOLUTIONS
20.13 Using the maximum principle and the hypothesis on g1andg2, for
all (x;y)2
[@
we have
min
(x;y)2@
u1(x;y) = min
(x;y)2@
g1(x;y)
u1(x;y)max
(x;y)2@
u1(x;y)
= max
(x;y)2@
g1(x;y)<max
(x;y)2@
g2(x;y)
min
(x;y)2@
g1(x;y) = min
(x;y)2@
u2(x;y)
u2(x;y)max
(x;y)2@
u2(x;y) = max
(x;y)2@
g2(x;y):
20.14 We have
(rncos (n)) =@2
@r2(rncos (n)) +1
r@
@r(rncos (n)) +1
r2@2
@2(rncos (n))
=n(n 1)rn 2cos (n) +nrn 2cos (n) rn 2n2cos (n) = 0
Likewise, ( rnsin (n)) = 0:
20.15u(r;) =1
2 r2
2a2cos 2:
20.16u(r;) = ln 2 + 4 a
r3cos 3:
Section 21
21.1 Convergent.
21.2 Divergent.
21.3 Convergent.
21.41
s 3; s> 3:
21.51
s2 5
s; s> 0:
21.6f(t) =e(t 1)2does not have a Laplace transform.
271
21.74
s 4
s2+2
s3; s> 0:
21.8e s
s2; s> 0:
21.9 e 2s
s+1
s2(e s e 2s); s6= 0:
21.10 tne st
s+n
sR
tn 1e stdt; s> 0:
21.11 (a) 0 (b) 0.
21.125
s+7+1
s2+2
s 2; s> 2:
21.13 3e2t; t0:
21.14 2t+e t; t0:
21.15 2(e 2t+e2t); t0:
21.162
s 1+5
s; s> 1:
21.17e s
s 3; s> 3:
21.181
2
1
s s2
s2+4!2
; s> 0:
21.193
s2+26; s> 0:
21.20s 3
(s 3)2+9; s> 3:
21.212
(s 4)3+3
(s 4)2+5
s 4; s> 4:
21.22 2 sin 5t+ 4e3t; t0:
21.235
6e3tt3; t0:
21.24
0; 0t<2
e9(t 2); t2:
272 ANSWERS AND SOLUTIONS
21.25 3e3t 3e t; t0:
21.26 4[e3(t 5) e 3(t 5)]H(t 5); t0:
21.27y(t) = 2e 4t+3[H(t 1) H(t 3)] 3[e 4(t 1)H(t 1) e 4(t 3)H(t
3)]; t0:
21.281
5e3t+1
20e 2t 1
4e2t; t0:
21.29et e 2t
3:
21.30t
2sint:
21.31t5
120:
21.321
2 e t+1
2e 2t:
21.33 t+et
2 e t
2:
Section 22
22.1u(x;t) = sin (x t) H(t x) sin (x t):
22.2u(x;t) = [sin (x t) H(t x) sin (x t)]e t:
22.3u(x;t) = 2e 42tsinx+ 6e 162tsin 2x:
22.4u(x;t) = [sin (x t) H(t x) sin (x t)]et:
22.5u(x;t) =t2e x te x+t:
22.6u(x;t) =
t 1
2x2
H
t 1
2x2
:
22.7u(x;t) =L 1
e s
cx
s2+1
=H
t x
c
sin
t x
c
:
22.8u(x;t) = 2 sinxcos 3t:
273
22.9u(x;y) =y(x+ 1) + 1:
22.10u(x;t) =L 1
e s
cx
s2+1
=h
t x
c
sin
t x
c
:
22.11u(x;t) =e 5xe 4tH(t):
22.12u(x;t) =L 1
T
se ps
cx+T
s
:
22.13u(x;t) = 5e 32tsin (x):
22.14u(x;t) = 40e tcosx
2:
22.15u(x;t) = 3 sinxcos 2t:
Section 23
23.1( 1)ni
n:
23.2f(x) =1
2+P1
n=11
nsin n
2
(einx+e inx):
23.3f(x) =sinha
P1
n= 1( 1)n(a+in)
(a2+n2)einx:
23.4f(x) =eix e ix
2i:
23.5f(x) =1
2
T+P 1
n= 1i
n[e int 1]eint+P1
n=1i
n[e int 1]eint
:
23.6 (a)f(x) =2
3+P 1
n= 12
n2( 1)neinx+P1
n=12
n2( 1)neinx:
(b)f(x) =2
3+P1
n=14
n2( 1)ncosnx:
274 ANSWERS AND SOLUTIONS
23.7 (a)
a0=2Z 1
2
1
2sinnxdx = 2
[cos
2 cos
2] = 0
an=2Z 1
2
1
2sinnxcos 2nxdx = 0
bn=2Z 1
2
1
2sinnxsin 2nxdx =8( 1)nn
4n2
(b)f(x) =4
P1
n= 1( 1)nn
i(1 4n2)e2nix:
23.8 (a)
a0=1
2Z2
2(2 x)dx= 4
an=1
2Z2
2(2 x) cosn
2x
dx= 0
bn=1
2Z2
2(2 x) sinn
2x
dx=4( 1)n
n
(b)f(x) = 2 +P 1
n= 12( 1)n+1i
ne(in
2x)+P1
n=12( 1)n+1i
ne(in
2x):
23.9an=cn+c n= 0:We have forjnjoddbn=i4
in=4
nand for
jnjevenbn= 0:
23.10 Note that for any complex number zwe havez+z= 2Re(z) and
z z= 2iRe(z):Thus,
cn+cn=an
which means that an= 2Re(cn):Likewise, we have
cn cn=ibn
That isibn= 2iIm(cn):Hence,bn= 2Im(cn):
23.11an= 2Re(cn) =1
nsin (nT) andbn=1 cos (nT)
n:
275
23.12f(x) =iP1
n= 1isin (2 in)
2 inein
2x:
23.13 (a) We have
f(t) =1 0<t< 1
0 1<t< 2
andf(t+ 2) =f(t) for allt2R:
(b) We have
a0=2
LZL
0f(x)dx=Z2
0dx=Z1
0dx= 1
an=Z1
0cosnxdx =sinn
n= 0:
(c) We have
bn=Z1
0sinnxdx =1 cosn
n=1 ( 1)n
n:
Hence,
bn=2
nifnis odd
0 ifnis even
(d) We have c0=a0
2=1
2and forn2Nwe have
cn=an ibn
2= i
nifnis odd
0 ifnis even
23.14 sin 3x=1
2(e3ix e 3ix):
23.15e ins
1 e inh
2in
:
Section 24
24.1
^f() =2sin
if6= 0
2 if= 0:
24.2@^u
@t+ic^u= 0
276 ANSWERS AND SOLUTIONS
^u(;0) = ^f():
24.3
@2^u
@t2= c22^u
^u(;0) = ^f()
^ut(;0) = ^g():
24.4
^uyy=2^u
^u(;0) = 0;^u(;L) =2 sina
:
24.51
i+1
+i=2
2+2:
24.6 We have
F[e xH(x)] =Z1
1e xH(x)e ixdx
=Z1
0e x(1+i)dx= e x(1+i)
1 +i 1
0=1
1 +i:
24.7 Using the duality property, we have
F1
1 +ix
=F[F[e H()]] = 2eH( ):
24.8 We have
F[f(x )] =Z1
1f(x )e ixdx
=e i Z1
1f(u)e iudu
=e i ^f()
277
whereu=x :
24.9 We have
F[ei xf(x)] =Z1
1ei xf(x)e ixdx=Z1
1eix( f(x)e ixdx=^f( ):
24.10 We will just prove the rst one. We have
F[cos ( x)f(x)] =F[f(x)ei x
2+f(x)e i x
2
=1
2[F[f(x)ei x] +F[f(x)e i x]]
=1
2[^f(x ) +^f(x+ )]:
24.11 Using the de nition and integration by parts we nd
F[f0(x)] =Z1
1f0(x)e ixdx
=f(x)e ix 1
1+ (i)Z1
1f(x)e ixdx
=f(x) cosx if(x) sinx+ (i)^f() = (i)^f()
where we used the fact that lim x!1f(x) = 0:
24.122
2(1 cos):
24.132
i(1 cosa):
24.14F 1[^f()] =1p
2e x2
2:
24.15F 1
1
a+i
=e ax; x0:
Section 25
25.1u(x;t) =f(x)F 1[ 1
jje jjy]:
278 ANSWERS AND SOLUTIONS
25.2u(x;t) =F 1[u(;t)] =e (x ct)2
4:
25.3
u(x;t) =r
4e tF 1[e 2(kt+
4)]
=r
4e tr
kt+
=4e x2
4(kt+
=4)
=p
4kt+
e x2
4kt+
e t:
25.4u(x;t) =1p
4ktR1
0e (x s)2
4ktds:
25.5
u(x;t) =etF 1[e 2t]
=e t1p
4te x2
4t:
25.6 We have
Z1
1e jjyeixd=Z0
1eyeixd+Z1
0e yeixd
=1
y+ixe(y+ix) 0
1+1
y+ixe( y+ix) 1
0
=1
y+ix+1
y+ix=2y
x2+y2:
25.7
u(x;y) =1
2Z1
1^f()e jjyeixd
=1
2f(x)2y
x2+y2
=1
2Z1
1f(x)2y
(x )2+y2d:
279
25.8 ^utt+ ( + )^ut+ ^u= c22^u:
25.9u(x;t) =e (x 3t):
25.10u(x;t) =e (x kt):
25.11u(x;t) =1p
4ktR1
1e s2 (x s)2
4ktds:
25.12u(x;t) = (x ct)2:
25.13u(x;t) =f(x)F 1[ 1
jje jjy]: